This deck is about structure-preserving maps between groups. It gives the homomorphism equation and shows why the identity and inverses are preserved automatically, then covers the image and the kernel and why the kernel is always a normal subgroup. It treats isomorphism as sameness up to relabeling, builds the quotient group and explains why coset multiplication needs normality, and closes with the First Isomorphism Theorem as the master factorization. It targets the real traps: assuming that every subgroup is normal, thinking that coset multiplication is always well defined, confusing the image with the codomain, and treating an injective homomorphism as automatically surjective.
Subject: Foundations of Higher Mathematics · 108 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. State the homomorphism condition and check whether a given map is a homomorphism.
2. Prove that a homomorphism preserves the identity and inverses, and compute its image and kernel.
3. Explain why the kernel is always a normal subgroup, and test whether a subgroup is normal.
4. Build a quotient group and say precisely why coset multiplication requires normality.
5. Apply the First Isomorphism Theorem to identify a quotient with a familiar group.
Warm-up
Discussion prompt
Before we open Group Homomorphisms & Quotient Groups: without looking back, what was the main idea of Subgroups, Cosets & Lagrange's Theorem, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers subgroups and the one-step test, cyclic subgroups and the order of an element, left cosets as a partition into equal-sized tiles, and the index of a subgroup. That leads to Lagrange's theorem and its corollaries: the order of an element divides the order of the group, every group of prime order is cyclic, and Fermat's little theorem follows. It targets the traps of assuming that a closed subset is automatically a subgroup, that left and right cosets always coincide, and that the converse of Lagrange's theorem holds.
Concept
A homomorphism is a map between two groups that respects their operations. Multiply first and then map, or map first and then multiply: you land in the same place.
\[ \varphi(ab) = \varphi(a)\,\varphi(b) \]
group homomorphism — A function from a group G to a group H such that the image of a product equals the product of the images, for all elements of G.
Counterexample
Discussion prompt
A homomorphism is a map between two groups that respects their operations. Multiply first and then map, or map first and then multiply: you land in the same place.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Concept
The subtlety hides in plain sight. On the left the product happens inside the source group; on the right the product happens inside the target group. One equation, two operations.
\[ \varphi:\; G \longrightarrow H, \qquad \varphi(a \ast_G b) = \varphi(a) \ast_H \varphi(b) \]
This is why a homomorphism can look nontrivial: it translates the source group's way of combining things into the target group's way of combining things.
Analogy
Discussion prompt
Explain The equation lives in two different groups by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
The subtlety hides in plain sight. On the left the product happens inside the source group; on the right the product happens inside the target group. One equation, two operations.
Intuition
Picture two groups as two languages. A homomorphism is a dictionary that translates elements from one to the other so that grammar is preserved: the translation of a combined phrase is the combination of the translations.
It need not be a perfect dictionary. Several source words can translate to the same target word, and some target words may never be used. What must never break is that combining is respected.
This single rule forces a surprising amount of structure for free, as the next few slides show.
Ranking
Put in order
Put the moves of Worked example: the exponential is a homomorphism into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The source operation is addition, so the input product is a sum; the target operation is multiplication.
Worked example
Take the additive real numbers as the source and the positive real numbers under multiplication as the target. Claim: the exponential map is a homomorphism.
\[ \varphi:\; (\mathbb{R}, +) \longrightarrow (\mathbb{R}_{>0}, \times), \qquad \varphi(x) = e^{x} \]
Write out both sides of the homomorphism equation
Why: The source operation is addition, so the input product is a sum; the target operation is multiplication.
\[ \varphi(x + y) = e^{x+y}, \qquad \varphi(x)\,\varphi(y) = e^{x} e^{y} \]
Use the exponent law to match them
Why: The identity that a sum of exponents is a product of powers is exactly the homomorphism condition here.
\[ e^{x+y} = e^{x} e^{y} \]
Verify on a concrete pair
Why: Test the boundary and one ordinary case: x equals 0 gives 1, the target identity; and a sample sum confirms the law numerically.
\[ \varphi(0) = e^{0} = 1, \qquad \varphi(1+2) = e^{3} = e^{1} e^{2} = \varphi(1)\varphi(2) \]
Picture it
Animation
Shows: Each line of the worked example "the exponential is a homomorphism", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Test the boundary and one ordinary case: x equals 0 gives 1, the target identity; and a sample sum confirms the law numerically.
Concept
You never have to assume this: it is forced. The source identity must map to the target identity.
\[ \varphi(e_G) = e_H \]
The reason is that the source identity is idempotent under the operation, and a homomorphism carries that fact across to the target, where only the target identity is idempotent.
Explain it
Discussion prompt
Explain A homomorphism sends the identity to the identity to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
You never have to assume this: it is forced. The source identity must map to the target identity.
Concept
Likewise, the image of an inverse is the inverse of the image. Undoing in the source becomes undoing in the target.
\[ \varphi(a^{-1}) = \varphi(a)^{-1} \]
Together with identity-preservation, this means a homomorphism carries the entire group skeleton across, not just the multiplication table.
Step zero
Discussion prompt
Worked example: proving identity and inverse are preserved — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Apply the map to the equation for the source identity
Answer:
Worked example
We derive both preservation facts from the single homomorphism equation, using nothing else.
Apply the map to the equation for the source identity
Why: The source identity satisfies e times e equals e; feed that through the homomorphism.
\[ \varphi(e_G) = \varphi(e_G e_G) = \varphi(e_G)\,\varphi(e_G) \]
Cancel one copy in the target group
Why: Multiply both sides by the inverse of that image in H; cancellation is legal because H is a group.
\[ e_H = \varphi(e_G) \]
Now map a times its inverse
Why: The source says a times a inverse is the identity; apply the map and use identity-preservation just proved.
\[ \varphi(a)\,\varphi(a^{-1}) = \varphi(a a^{-1}) = \varphi(e_G) = e_H \]
Verify the inverse claim
Why: The last line says the image of a inverse is a right inverse of the image of a in H; by uniqueness of inverses in a group it is the inverse, confirming the identity for a sample element as well.
\[ \varphi(a^{-1}) = \varphi(a)^{-1} \]
Picture it
Animation
Shows: Each line of the worked example "proving identity and inverse are preserved", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The last line says the image of a inverse is a right inverse of the image of a in H; by uniqueness of inverses in a group it is the inverse, confirming the identity for a sample element as well.
Concept
Because powers are preserved, a homomorphism can only shrink or preserve the order of an element, never enlarge it.
\[ a^{m} = e_G \;\Longrightarrow\; \varphi(a)^{m} = \varphi(a^{m}) = \varphi(e_G) = e_H \]
So the order of the image divides the order of the element. This is a fast sanity check: an element of order three can never map to an element of order two.
Estimation
Predict first
Let the source be the symmetric group on three symbols and the target be the two-element multiplicative group of plus and minus one. Send each permutation to its sign.
Commit before you compute: what does Worked example: the sign homomorphism on permutations come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the multiplication of signs
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Multiplying two odd permutations lands in the even class, and minus one times minus one is plus one, matching the computed product exactly.
Worked example
Let the source be the symmetric group on three symbols and the target be the two-element multiplicative group of plus and minus one. Send each permutation to its sign.
\[ \operatorname{sgn}:\; S_3 \longrightarrow \{+1, -1\}, \qquad \operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau) \]
Record the sign of each element type
Why: The identity and the two three-cycles are even; the three transpositions are odd.
\[ \operatorname{sgn}(e) = \operatorname{sgn}(123) = \operatorname{sgn}(132) = +1, \quad \operatorname{sgn}(12) = \operatorname{sgn}(13) = \operatorname{sgn}(23) = -1 \]
Check the homomorphism equation on a product of two transpositions
Why: A product of two transpositions is a three-cycle, which is even; the signs multiply to give plus one.
\[ (12)(13) = (132), \qquad \operatorname{sgn}(12)\operatorname{sgn}(13) = (-1)(-1) = +1 = \operatorname{sgn}(132) \]
Verify against the multiplication of signs
Why: Multiplying two odd permutations lands in the even class, and minus one times minus one is plus one, matching the computed product exactly.
\[ \operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau) \;\text{holds on the tested product} \]
Picture it
Animation
Shows: Each line of the worked example "the sign homomorphism on permutations", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A product of two transpositions is a three-cycle, which is even; the signs multiply to give plus one.
Trap
Tempting claim: since the exponential map is written with the multiplicative real numbers as its target, its image is all nonzero reals.
\[ \text{Wrong: } \operatorname{im}(\exp) = \mathbb{R}^{\times} \]
But the exponential never outputs a negative number, so the negative reals are in the codomain yet never hit. The codomain is where outputs are allowed to live; the image is where they actually land.
The image is the set of values actually achieved, which here is only the positive reals.
\[ \operatorname{im}(\exp) = \mathbb{R}_{>0} \subsetneq \mathbb{R}^{\times} \]
Surjectivity is a claim about the codomain, so always name the codomain and the image separately. They coincide only when the map is onto.
Concept
The image of a homomorphism is the set of outputs it produces, sitting inside the target group.
\[ \operatorname{im}\varphi = \{\, \varphi(g) : g \in G \,\} \subseteq H \]
It is never empty, because at the very least the identity is hit. The question of interest is how much of H it fills.
Concept
The image is not just any subset: it is itself a group under the target operation.
\[ \operatorname{im}\varphi \le H \]
This is because a homomorphism carries the source group structure faithfully onto its outputs, so the outputs are closed, contain the identity, and are closed under inverses.
Missing information
Discussion prompt
We check the three subgroup requirements directly for the image, using only preservation facts.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The target identity equals the image of the source identity, so it lies in the image.
Worked example
We check the three subgroup requirements directly for the image, using only preservation facts.
Nonempty and contains the identity
Why: The target identity equals the image of the source identity, so it lies in the image.
\[ e_H = \varphi(e_G) \in \operatorname{im}\varphi \]
Closed under the target operation
Why: A product of two images is the image of a product, which is again an output.
\[ \varphi(a)\,\varphi(b) = \varphi(ab) \in \operatorname{im}\varphi \]
Closed under inverses
Why: The inverse of an image is the image of the inverse, again an output.
\[ \varphi(a)^{-1} = \varphi(a^{-1}) \in \operatorname{im}\varphi \]
Verify all three tests are met
Why: The image contains the identity, is closed under products, and is closed under inverses, so by the subgroup test it is a subgroup of H.
\[ \operatorname{im}\varphi \le H \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the image is a subgroup", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The image contains the identity, is closed under products, and is closed under inverses, so by the subgroup test it is a subgroup of H.
Fill the middle
Fill in the blanks
From Worked example: the determinant homomorphism — finish the line. Write what belongs on the right of the equals sign before you look.
\det:\; GL_2(\mathbb\det(A)\det(B)) \longrightarrow \mathbb___^___, \qquad \det(AB) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The multiplicativity of the determinant is exactly the homomorphism condition for this map.
Worked example
Let the source be the invertible two-by-two real matrices under multiplication and the target be the nonzero reals under multiplication. Send a matrix to its determinant.
\[ \det:\; GL_2(\mathbb{R}) \longrightarrow \mathbb{R}^{\times}, \qquad \det(AB) = \det(A)\det(B) \]
Confirm the homomorphism equation
Why: The multiplicativity of the determinant is exactly the homomorphism condition for this map.
\[ \det(AB) = \det(A)\det(B) \]
Identify the image
Why: Every nonzero real r is achieved, for example by a diagonal matrix with entries r and 1, so the map is onto.
\[ \det\!\begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix} = r, \qquad \operatorname{im}(\det) = \mathbb{R}^{\times} \]
Verify surjectivity on a sample value
Why: Picking r equal to minus five exhibits a specific invertible matrix whose determinant is minus five, confirming the negative reals are hit, unlike the exponential example.
\[ \det\!\begin{pmatrix} -5 & 0 \\ 0 & 1 \end{pmatrix} = -5 \in \operatorname{im}(\det) \]
Concept
The kernel is the set of source elements that the homomorphism crushes down to the target identity.
\[ \ker\varphi = \{\, g \in G : \varphi(g) = e_H \,\} \subseteq G \]
kernel — The preimage of the target identity under a homomorphism: all source elements sent to the identity of the target group.
Intuition
If the image is what survives translation, the kernel is what gets flattened. A large kernel means the map forgets a lot; a trivial kernel means it forgets nothing.
Figure (svg): Several source elements on the left all mapping by arrows onto a single identity point on the right, with the collapsing elements grouped as the kernel.
The genius of the kernel is that it turns the abstract question, how far is this map from injective, into a concrete subgroup you can compute with.
Concept
Just like the image, the kernel is a group in its own right, this time inside the source.
\[ \ker\varphi \le G \]
It contains the identity, and if two elements both map to the target identity, so do their product and their inverses. In a moment we will see the kernel is even better than a subgroup.
Concept
Two elements share the same output exactly when they differ by something in the kernel. So the set of elements mapping to a given value is a translate of the kernel.
\[ \varphi(x) = \varphi(y) \iff x^{-1}y \in \ker\varphi \iff y \in x\,\ker\varphi \]
Every fiber is one coset of the kernel, all fibers have the same size, and this is the seed of the quotient construction later in the deck.
Concept
Because the fibers are cosets of the kernel, the map separates points precisely when the kernel is as small as possible.
\[ \varphi \text{ is injective} \iff \ker\varphi = \{e_G\} \]
This replaces a check over all pairs of inputs with a single computation: find the kernel. It is one of the most used facts in the whole subject.
Step zero
Discussion prompt
Worked example: injective iff kernel trivial — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Forward: assume injective and take any kernel element
Answer:
Worked example
We prove both directions of the equivalence from the definitions alone.
Forward: assume injective and take any kernel element
Why: A kernel element maps to the identity, and so does the source identity; injectivity forces them equal.
\[ \varphi(a) = e_H = \varphi(e_G) \;\Rightarrow\; a = e_G \]
Backward: assume trivial kernel and two elements with equal image
Why: Combine the images so the equation lands on the identity, then convert with inverse-preservation.
\[ \varphi(a) = \varphi(b) \;\Rightarrow\; \varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} = e_H \]
Use the trivial kernel to conclude equality
Why: The element ab inverse is in the kernel, which is just the identity, so a equals b.
\[ ab^{-1} \in \ker\varphi = \{e_G\} \;\Rightarrow\; ab^{-1} = e_G \;\Rightarrow\; a = b \]
Verify both directions close the equivalence
Why: Injective gives trivial kernel and trivial kernel gives injective, so the two conditions are genuinely the same, confirmed on the general elements used above.
\[ \varphi \text{ injective} \iff \ker\varphi = \{e_G\} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "injective iff kernel trivial", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Injective gives trivial kernel and trivial kernel gives injective, so the two conditions are genuinely the same, confirmed on the general elements used above.
Fill the middle
Fill in the blanks
From Worked example: the kernel of the determinant — finish the line. Write what belongs on the right of the equals sign before you look.
\ker(\det) = SL_n(\mathbb{R})
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The target identity of the nonzero reals under multiplication is one, so the kernel is the determinant-one matrices.
Worked example
Return to the determinant map on invertible matrices. Its kernel is the set of matrices sent to the target identity, which is the number one.
Write the kernel condition
Why: The target identity of the nonzero reals under multiplication is one, so the kernel is the determinant-one matrices.
\[ \ker(\det) = \{\, A \in GL_n(\mathbb{R}) : \det A = 1 \,\} \]
Name this subgroup
Why: The determinant-one matrices form the special linear group, a standard and important subgroup.
\[ \ker(\det) = SL_n(\mathbb{R}) \]
Verify with a specific membership test
Why: A shear matrix has determinant one, so it lies in the kernel, while a scaling matrix of determinant two does not, confirming the kernel is exactly the determinant-one matrices.
\[ \det\!\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = 1 \in \ker, \qquad \det\!\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} = 2 \notin \ker \]
Picture it
Animation
Shows: Each line of the worked example "the kernel of the determinant", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A shear matrix has determinant one, so it lies in the kernel, while a scaling matrix of determinant two does not, confirming the kernel is exactly the determinant-one matrices.
Trap
Tempting claim: a homomorphism with trivial kernel is an isomorphism, since trivial kernel means injective.
\[ \text{Wrong: } \ker\varphi = \{e_G\} \;\Rightarrow\; \varphi \text{ is an isomorphism} \]
Doubling on the integers is injective, its kernel is just zero, yet it never outputs an odd number. Injective was only half of what an isomorphism needs.
Trivial kernel gives injective; you still owe a separate proof of surjectivity onto the stated codomain.
\[ \varphi:\mathbb{Z}\to\mathbb{Z}, \; \varphi(n)=2n: \quad \ker\varphi=\{0\}, \;\; \operatorname{im}\varphi = 2\mathbb{Z} \subsetneq \mathbb{Z} \]
In the special case of finite groups of equal size, injective does force surjective by counting; in general it does not. Never conflate the two halves.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Trivial kernel gives injective; you still owe a separate proof of surjectivity onto the stated codomain.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
To conjugate one element by another is to sandwich it between that element and its inverse. Think of it as viewing the first element from the second one's point of view.
\[ a \longmapsto g a g^{-1} \]
Conjugation preserves all structural features such as order, so it is a symmetry of the group itself. A subgroup that is unchanged by every conjugation will be the key player next.
Concept
A subgroup is normal when conjugating any of its elements by any group element keeps you inside the subgroup.
\[ N \trianglelefteq G \iff gNg^{-1} = N \;\text{ for all } g \in G \]
normal subgroup — A subgroup N of G such that conjugation by every element of G maps N onto itself; equivalently, its left and right cosets coincide.
Definition probe
Sort into buckets
Every line below is part of the definition of kernel or of normal subgroup — one or the other, never both. Put each where it belongs.
Intuition
Conjugation invariance has a clean cosets translation: sliding the subgroup on the left gives the same set as sliding it on the right.
\[ gN = Ng \;\text{ for all } g \in G \]
This is the property that will let cosets be multiplied unambiguously. Normality is precisely the condition that makes the coset arithmetic of the next section legal.
Hypothesis
Predict first
Worked example: normal and non-normal subgroups of S three is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Test the transposition subgroup by conjugation
Why: Conjugating the transposition swapping one and two by the transposition swapping one and three relabels the symbols and produces a different transposition.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
In the symmetric group on three symbols, compare a two-element subgroup generated by a transposition with the three-element alternating subgroup.
Test the transposition subgroup by conjugation
Why: Conjugating the transposition swapping one and two by the transposition swapping one and three relabels the symbols and produces a different transposition.
\[ (13)(12)(13)^{-1} = (23) \notin \{\, e, (12) \,\} \]
Conclude this subgroup is not normal
Why: A single conjugation escaped the subgroup, so the conjugation-invariance condition fails.
\[ \langle (12) \rangle \;\text{ is not normal in } S_3 \]
Test the alternating subgroup
Why: The alternating subgroup has index two, and any index-two subgroup is automatically normal because there are only two cosets on each side.
\[ [S_3 : A_3] = 2 \;\Rightarrow\; A_3 \trianglelefteq S_3 \]
Verify the contrast on conjugation
Why: Conjugating a three-cycle in the alternating subgroup gives another three-cycle, which stays inside, confirming normality against the failed transposition case.
\[ (12)(123)(12)^{-1} = (132) \in A_3 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "normal and non-normal subgroups of S three", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Conjugating a three-cycle in the alternating subgroup gives another three-cycle, which stays inside, confirming normality against the failed transposition case.
Concept
Kernels are special among subgroups: they are automatically closed under conjugation, no matter how twisted the group is.
\[ \ker\varphi \trianglelefteq G \]
This is the deep reason kernels matter. In fact every normal subgroup is the kernel of some homomorphism, so kernels and normal subgroups are two names for the same thing.
Ranking
Put in order
Put the moves of Worked example: proving the kernel is normal into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Split across the product using the homomorphism property three times.
Worked example
Take any kernel element and any group element, and show the conjugate is still in the kernel.
Apply the map to the conjugate
Why: Split across the product using the homomorphism property three times.
\[ \varphi(gkg^{-1}) = \varphi(g)\,\varphi(k)\,\varphi(g)^{-1} \]
Use that k is in the kernel
Why: The middle factor is the target identity, so the outer factors cancel.
\[ = \varphi(g)\,e_H\,\varphi(g)^{-1} = e_H \]
Conclude the conjugate lies in the kernel
Why: The conjugate maps to the identity, so by definition it is a kernel element, for every choice of g and k.
\[ gkg^{-1} \in \ker\varphi \;\text{ for all } g \in G, \; k \in \ker\varphi \]
Verify normality
Why: Conjugation by any element keeps the kernel inside itself, which is exactly the normality condition, confirmed for arbitrary g and k.
\[ g\,\ker\varphi\,g^{-1} \subseteq \ker\varphi \;\Rightarrow\; \ker\varphi \trianglelefteq G \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "proving the kernel is normal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Conjugation by any element keeps the kernel inside itself, which is exactly the normality condition, confirmed for arbitrary g and k.
Trap
Tempting claim: any subgroup is normal, so we can always form cosets that multiply, and we may as well quotient by anything.
\[ \text{Wrong: every } H \le G \text{ is normal} \]
This feels true because it holds in every abelian group, where left and right cosets always agree. Strong students who cut their teeth on the integers absorb it as a reflex.
In a non-abelian group normality can genuinely fail, as the transposition subgroup of S three showed.
\[ \langle (12) \rangle \le S_3, \quad (13)\langle(12)\rangle \ne \langle(12)\rangle(13) \]
Only when the group is abelian is every subgroup normal. Otherwise you must check conjugation before you are allowed to build a quotient.
Concept
When a subgroup is normal, its cosets can be treated as the elements of a brand new group. You multiply two cosets by multiplying any representatives and taking the coset of the result.
\[ (aN)(bN) = (ab)N \]
quotient group — The set of cosets of a normal subgroup N in G, with the operation that multiplies cosets by multiplying representatives; written G modulo N.
Intuition
Forming the quotient is a controlled act of forgetting: you decree that everything in the normal subgroup counts as the identity, and then you see what group structure remains.
\[ N \rightsquigarrow e \;\text{ in } G/N \]
This is the exact same move as reducing integers modulo n: you glue together numbers that differ by a multiple of n, and arithmetic survives the gluing.
Concept
The coset product rule names representatives, but a coset has many. For the rule to define an operation, different representatives must give the same answer.
\[ aN = a'N, \; bN = b'N \;\overset{?}{\Longrightarrow}\; (ab)N = (a'b')N \]
This well-definedness question is the whole ballgame. It succeeds exactly when the subgroup is normal, and it fails otherwise.
Estimation
Predict first
Assume the subgroup is normal and show that swapping to any other representatives leaves the product coset unchanged.
Commit before you compute: what does Worked example: coset multiplication is well-defined when N… come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the product coset is unchanged
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The alternate product lies in the same coset as the original product, so the operation does not depend on the chosen representatives; it is well-defined.
Worked example
Assume the subgroup is normal and show that swapping to any other representatives leaves the product coset unchanged.
Write the alternate representatives
Why: Same coset means the new representative differs from the old by a subgroup element.
\[ a' = an, \quad b' = bm, \qquad n, m \in N \]
Expand the alternate product and insert a strategic factor
Why: Slide a copy of b past n by inserting b inverse times b, so normality can act on the middle.
\[ a'b' = a n b m = ab\,(b^{-1} n b)\,m \]
Apply normality to absorb the middle into N
Why: Normality says the conjugate of a subgroup element is again a subgroup element, so the whole tail lands in N.
\[ b^{-1} n b \in N \;\Rightarrow\; (b^{-1} n b)\,m \in N \;\Rightarrow\; a'b' \in (ab)N \]
Verify the product coset is unchanged
Why: The alternate product lies in the same coset as the original product, so the operation does not depend on the chosen representatives; it is well-defined.
\[ (a'b')N = (ab)N \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "coset multiplication is well-defined when N is normal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The alternate product lies in the same coset as the original product, so the operation does not depend on the chosen representatives; it is well-defined.
Trap
Tempting claim: cosets can always be multiplied by picking representatives, for any subgroup at all.
Test it on the non-normal transposition subgroup of the permutations of three symbols. One coset has two representatives; multiply each by the same element.
\[ (13)\cdot(13) = e \in H, \qquad (123)\cdot(13) = (23) \in (23)H \]
The two representatives of the one coset gave products in two different cosets, so the rule is not even a function. Without normality there is no quotient.
\[ H \ne (23)H \;\Rightarrow\; \text{the product depends on the representative} \]
This is exactly what the well-definedness proof needed normality for. The failure is not a technicality; it is the reason normal subgroups are singled out.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Worked example
Work inside the additive group of integers modulo six, which is abelian, so every subgroup is normal. Quotient by the subgroup generated by three.
\[ G = \mathbb{Z}/6\mathbb{Z}, \qquad N = \langle 3 \rangle = \{0, 3\} \]
List the cosets
Why: Add each residue to the subgroup; residues that differ by three land in the same coset.
\[ 0+N = \{0,3\}, \quad 1+N = \{1,4\}, \quad 2+N = \{2,5\} \]
Count and add the cosets
Why: There are three cosets, and adding the coset of one to itself steps through them, returning to the identity after three additions.
\[ (1+N)+(1+N) = 2+N, \qquad (2+N)+(1+N) = 3+N = 0+N \]
Verify the quotient is the integers mod three
Why: The quotient has three elements and is cyclic, generated by the coset of one with order exactly three, so it is isomorphic to the integers modulo three.
\[ (\mathbb{Z}/6\mathbb{Z})/\langle 3 \rangle \;\cong\; \mathbb{Z}/3\mathbb{Z} \]
Concept
It helps to see the cosets as a tiling. Each coset has the same size as the subgroup, and together they partition the whole group with no overlaps.
Figure (svg): The six elements of the integers mod six grouped into three boxes: zero and three, one and four, two and five, showing the coset partition.
The quotient group is the group of tiles. Each tile is one element of the quotient, and multiplying tiles is what coset multiplication computes.
Concept
An isomorphism is a homomorphism that is both injective and surjective. Its inverse is automatically a homomorphism too, so it perfectly matches the two groups.
\[ \varphi:\; G \xrightarrow{\;\cong\;} H \]
isomorphism — A bijective group homomorphism; when one exists the groups are called isomorphic and are regarded as the same group with relabeled elements.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of group homomorphism, kernel, normal subgroup, quotient group, isomorphism as Group Homomorphisms & Quotient Groups uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
If two groups are isomorphic, then anything you can say about one purely in terms of the operation is also true of the other. They differ only in the names printed on the elements.
So isomorphism is the algebraic version of the bijection idea from set theory: same size becomes same structure. It is the notion of sameness that group theory actually cares about.
This is why the goal of so many theorems is to prove some mystery group is isomorphic to a familiar one.
Concept
Because an isomorphism matches the operations exactly, it carries across all structural features: the size of the group, whether it is abelian, and the order of every element.
\[ \varphi \text{ iso} \;\Rightarrow\; |G| = |H|, \quad \text{order}(g) = \text{order}(\varphi(g)) \]
This gives a practical test for non-isomorphism: if two groups disagree on any such invariant, no isomorphism between them can exist.
Step zero
Discussion prompt
Worked example: two groups of order four that are not isomorphic — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the element orders in the cyclic group
Answer:
Worked example
Both the cyclic group of order four and the Klein four-group have four elements and are abelian. Yet they are not isomorphic. The order of elements tells them apart.
Find the element orders in the cyclic group
Why: A generator of the cyclic group of order four has order four; such an element exists.
\[ \mathbb{Z}/4\mathbb{Z}: \; \text{order of } 1 \text{ is } 4 \]
Find the element orders in the Klein four-group
Why: In the Klein four-group every non-identity element squares to the identity, so the largest order present is two.
\[ V = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}: \; \text{every nonidentity element has order } 2 \]
Verify no isomorphism can exist
Why: An isomorphism preserves order, but one group has an element of order four and the other has none, so no bijective homomorphism can match them; they are genuinely different groups.
\[ \mathbb{Z}/4\mathbb{Z} \;\not\cong\; \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \]
Picture it
Animation
Shows: Each line of the worked example "two groups of order four that are not isomorphic", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: An isomorphism preserves order, but one group has an element of order four and the other has none, so no bijective homomorphism can match them; they are genuinely different groups.
Concept
Homomorphisms chain: if you map from the first group to the second and then from the second to the third, the combined map is again structure-preserving.
\[ (\psi \circ \varphi)(ab) = \psi(\varphi(ab)) = \psi(\varphi(a)\varphi(b)) = (\psi\circ\varphi)(a)\,(\psi\circ\varphi)(b) \]
So groups and homomorphisms form a world you can navigate by composition, which is what makes them a category. Isomorphisms are exactly the invertible arrows in that world.
Worked example
The exponential and the natural logarithm are inverse homomorphisms between the additive reals and the positive reals under multiplication.
\[ \varphi(x) = e^{x}, \qquad \psi(t) = \ln t \]
Confirm the logarithm is a homomorphism
Why: The logarithm of a product is the sum of logarithms, which is the homomorphism condition from multiplication to addition.
\[ \psi(st) = \ln(st) = \ln s + \ln t = \psi(s) + \psi(t) \]
Compose the two maps
Why: Applying exponential then logarithm returns the original input, since the two are inverse functions.
\[ (\psi \circ \varphi)(x) = \ln(e^{x}) = x \]
Verify the composite is the identity homomorphism
Why: The composite fixes every element and respects addition, so exponential is an isomorphism with logarithm as its inverse, checked on a sample input as well.
\[ (\psi\circ\varphi)(2) = \ln(e^{2}) = 2 \;\checkmark \]
Concept
Here is the master theorem of the whole topic. Every homomorphism, once you collapse its kernel, becomes an isomorphism onto its image.
\[ G/\ker\varphi \;\cong\; \operatorname{im}\varphi \]
The kernel measures exactly what the map forgets, so quotienting by it removes the forgetting and leaves a perfect copy of the image. Nothing is lost and nothing is doubled.
Intuition
The theorem says any homomorphism can be read in two clean stages. First collapse the kernel by projecting onto cosets; then relabel those cosets bijectively onto the image.
Figure (svg): A triangle diagram: G maps by phi to H across the top; G projects down to the quotient G over kernel; the quotient maps by an isomorphism onto the image inside H.
The first stage loses information, the second loses none. This is the same bookkeeping shape as the rank plus nullity theorem in linear algebra and the Lindenbaum quotient in logic.
Concept
The bridge from a normal subgroup to the theorem is the projection that sends each element to its own coset.
\[ \pi:\; G \longrightarrow G/N, \qquad \pi(g) = gN \]
This map is a surjective homomorphism, and its kernel is exactly the subgroup you quotiented by. So every normal subgroup really is a kernel, closing the loop.
Missing information
Discussion prompt
Check that the canonical projection preserves the operation and pin down its kernel.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The coset of a product is the product of cosets, which is precisely the quotient operation.
Worked example
Check that the canonical projection preserves the operation and pin down its kernel.
Confirm the homomorphism property
Why: The coset of a product is the product of cosets, which is precisely the quotient operation.
\[ \pi(ab) = (ab)N = (aN)(bN) = \pi(a)\,\pi(b) \]
Compute the kernel
Why: An element maps to the identity coset exactly when its coset equals N, which happens precisely when the element is in N.
\[ \pi(g) = N \iff gN = N \iff g \in N \]
Verify the kernel is N and the map is onto
Why: The kernel is exactly N and every coset is hit by any of its representatives, so the projection is a surjective homomorphism with kernel N, as claimed.
\[ \ker\pi = N, \qquad \operatorname{im}\pi = G/N \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the projection is a homomorphism with kernel N", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The kernel is exactly N and every coset is hit by any of its representatives, so the projection is a surjective homomorphism with kernel N, as claimed.
Estimation
Predict first
Let the map reduce each integer to its remainder modulo n. This is the cleanest instance of the theorem.
Commit before you compute: what does Worked example: the theorem applied to reduction modulo n come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify on a sample modulus
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. With n equal to five the kernel is the multiples of five and the quotient has exactly the five residue classes, matching the group of integers modulo five.
Worked example
Let the map reduce each integer to its remainder modulo n. This is the cleanest instance of the theorem.
\[ \varphi:\; \mathbb{Z} \longrightarrow \mathbb{Z}/n\mathbb{Z}, \qquad \varphi(k) = k \bmod n \]
Identify the image and the kernel
Why: Every residue is achieved, so the map is onto; an integer reduces to zero exactly when it is a multiple of n.
\[ \operatorname{im}\varphi = \mathbb{Z}/n\mathbb{Z}, \qquad \ker\varphi = n\mathbb{Z} \]
Feed both into the theorem
Why: The source modulo the kernel is isomorphic to the image, which turns the abstract quotient into the concrete group of residues.
\[ \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}/\ker\varphi \;\cong\; \operatorname{im}\varphi = \mathbb{Z}/n\mathbb{Z} \]
Verify on a sample modulus
Why: With n equal to five the kernel is the multiples of five and the quotient has exactly the five residue classes, matching the group of integers modulo five.
\[ \mathbb{Z}/5\mathbb{Z} \cong \{\,\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\,\} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the theorem applied to reduction modulo n", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: With n equal to five the kernel is the multiples of five and the quotient has exactly the five residue classes, matching the group of integers modulo five.
Step zero
Discussion prompt
Worked example: the theorem applied to the determinant — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Insert the pieces into the theorem
Answer:
Worked example
Apply the theorem to the determinant map, whose image and kernel we already computed.
\[ \det:\; GL_n(\mathbb{R}) \longrightarrow \mathbb{R}^{\times}, \qquad \operatorname{im}(\det) = \mathbb{R}^{\times}, \quad \ker(\det) = SL_n(\mathbb{R}) \]
Insert the pieces into the theorem
Why: Quotienting the invertible matrices by the determinant-one matrices leaves an isomorphic copy of the nonzero reals.
\[ GL_n(\mathbb{R})/SL_n(\mathbb{R}) \;\cong\; \mathbb{R}^{\times} \]
Read the meaning
Why: Two invertible matrices land in the same coset exactly when they share a determinant, so the coset is completely described by that single nonzero number.
\[ A\,SL_n(\mathbb{R}) \longleftrightarrow \det A \]
Verify the correspondence is a bijection onto the nonzero reals
Why: Each nonzero determinant value names exactly one coset and every coset has a determinant, so the assignment is a well-defined isomorphism, confirming the theorem here.
\[ GL_n(\mathbb{R})/SL_n(\mathbb{R}) \;\cong\; \mathbb{R}^{\times} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the theorem applied to the determinant", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each nonzero determinant value names exactly one coset and every coset has a determinant, so the assignment is a well-defined isomorphism, confirming the theorem here.
Ranking
Put in order
Put the moves of Worked example: the theorem applied to the sign map into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The kernel is the even permutations and the image is the two-element sign group, so quotienting gives that two-element group.
Worked example
Apply the theorem to the sign homomorphism on the permutations of three symbols.
\[ \operatorname{sgn}:\; S_3 \longrightarrow \{+1,-1\}, \qquad \operatorname{im} = \{+1,-1\}, \quad \ker = A_3 \]
Insert kernel and image
Why: The kernel is the even permutations and the image is the two-element sign group, so quotienting gives that two-element group.
\[ S_3/A_3 \;\cong\; \{+1,-1\} \;\cong\; \mathbb{Z}/2\mathbb{Z} \]
Check the sizes agree
Why: The permutation group has six elements and the alternating subgroup has three, so the quotient has two cosets, matching the order of the sign group.
\[ |S_3|/|A_3| = 6/3 = 2 = |\mathbb{Z}/2\mathbb{Z}| \]
Verify the two cosets are even and odd
Why: One coset is the even permutations and the other is the odd ones, and their product rule matches addition modulo two, confirming the isomorphism with the integers modulo two.
\[ \{\text{even}\}\cdot\{\text{odd}\} = \{\text{odd}\}, \quad \{\text{odd}\}\cdot\{\text{odd}\} = \{\text{even}\} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the theorem applied to the sign map", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The permutation group has six elements and the alternating subgroup has three, so the quotient has two cosets, matching the order of the sign group.
Concept
Two facts now meet in the middle. Every kernel is a normal subgroup, and every normal subgroup is the kernel of its own canonical projection.
\[ N \trianglelefteq G \iff N = \ker\pi \;\text{ for } \pi: G \to G/N \]
So there is nothing more general to search for. The normal subgroups are exactly the subgroups you can quotient by, and they are exactly the kernels. Three descriptions, one class of objects.
Explain it
Discussion prompt
Explain Normal subgroups and kernels are the same objects to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Two facts now meet in the middle. Every kernel is a normal subgroup, and every normal subgroup is the kernel of its own canonical projection.
Intuition
One habit to build: the theorem produces an isomorphism onto the image, never a claim about the codomain. Read the right-hand side as the values actually achieved.
When the map happens to be surjective the image fills the codomain and the two coincide, which is why the reduction and determinant examples felt so clean. When it does not, the image is the honest answer.
Keeping image and codomain separate is the same discipline that the earlier trap demanded; here it decides what group your quotient is isomorphic to.
Analogy
Discussion prompt
Explain Isomorphic to the image, not equal to the codomain by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
One habit to build: the theorem produces an isomorphism onto the image, never a claim about the codomain. Read the right-hand side as the values actually achieved.
Concept
Step back and notice the shape. A structure-preserving map has a piece it forgets and a piece it faithfully copies, and quotienting by the forgotten piece recovers the copy exactly.
You saw it with equivalence relations collapsing to a quotient set, you see it here with kernels and quotient groups, and you will see it again as ideals and quotient rings and as rank plus nullity for linear maps.
Structure-preserving map, kernel, quotient, isomorphism onto the image: memorize this quartet, because it is the reusable engine of abstract algebra and beyond.
Counterexample
Discussion prompt
Step back and notice the shape. A structure-preserving map has a piece it forgets and a piece it faithfully copies, and quotienting by the forgotten piece recovers the copy exactly.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Structure-preserving map, kernel, quotient, isomorphism onto the image: memorize this quartet, because it is the reusable engine of abstract algebra and beyond.
Pattern
1. Build or spot a homomorphism out of the group of interest
Why: Choose a target where the structure you want to detect is visible, such as the sign, the determinant, or a reduction map.
2. Compute the image
Why: Find which target elements are actually hit; this becomes the right-hand side of the isomorphism.
3. Compute the kernel
Why: Find everything sent to the identity; it is automatically a normal subgroup, so you are allowed to quotient by it.
4. Invoke the First Isomorphism Theorem
Why: The source modulo the kernel is isomorphic to the image, which names the quotient as a familiar group with no further work.
5. Sanity-check orders in the finite case
Why: The order of the group divided by the order of the kernel must equal the order of the image; if it does not, recheck the image or the kernel.
Real world
Discussion prompt
Outside this lesson: where does Group Homomorphisms & Quotient Groups actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe: use a homomorphism to identify a quotient is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck is about structure-preserving maps between groups. It gives the homomorphism equation and shows why the identity and inverses are preserved automatically, then covers the image and the kernel and why the kernel is always a normal subgroup. It treats isomorphism as sameness up to relabeling, builds the quotient group and explains why coset multiplication needs normality, and closes with the First Isomorphism Theorem as the master factorization. It targets the real traps: assuming that every subgroup is normal, thinking that coset multiplication is always well defined, confusing the image with the codomain, and treating an injective homomorphism as automatically surjective.
Elimination
Eliminate the wrong options
What is the kernel of this determinant homomorphism?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The kernel is the preimage of the target identity, which is the number one, so it is exactly the determinant-one matrices, namely the special linear group.
Check
The determinant map goes from the invertible two-by-two real matrices to the nonzero reals under multiplication.
Check your understanding
What is the kernel of this determinant homomorphism?
Answer: A
Why: The kernel is the preimage of the target identity, which is the number one, so it is exactly the determinant-one matrices, namely the special linear group.
Prediction
Predict first
Why must the subgroup N be normal before its cosets form a group?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: So that coset multiplication by representatives is well-defined
Why: Normality is exactly the condition that makes the coset product independent of which representatives are chosen, so the operation on cosets is a genuine function.
Check
You want the cosets of a subgroup to form a group under multiplication of representatives.
Check your understanding
Why must the subgroup N be normal before its cosets form a group?
Answer: A
Why: Normality is exactly the condition that makes the coset product independent of which representatives are chosen, so the operation on cosets is a genuine function.
Check
Let N be the two-element subgroup consisting of zero and four inside the additive integers modulo eight.
Check your understanding
Which group is the quotient of the integers mod eight by this subgroup isomorphic to?
Answer: A
Why: The subgroup has two elements, so the quotient has eight divided by two equals four cosets, and the coset of one has order four, making the quotient cyclic of order four.
Commit first
Predict first
By the First Isomorphism Theorem, the quotient of the integers by that kernel is isomorphic to which group?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: The integers modulo twelve
Why: The theorem identifies the source modulo the kernel with the image, and the reduction map is onto the integers modulo twelve, so the quotient is exactly that group.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Reduction sends each integer to its residue modulo twelve, onto the integers mod twelve, with kernel the multiples of twelve.
Check your understanding
By the First Isomorphism Theorem, the quotient of the integers by that kernel is isomorphic to which group?
Answer: A
Why: The theorem identifies the source modulo the kernel with the image, and the reduction map is onto the integers modulo twelve, so the quotient is exactly that group.
Prediction
Predict first
Which statement about the doubling map is correct?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: It is injective but not surjective
Why: Its kernel is only zero, so the map is injective, but the odd integers are never outputs, so it is not surjective onto the integers.
Check
The doubling map on the integers sends each n to two n, and it is a homomorphism under addition.
Check your understanding
Which statement about the doubling map is correct?
Answer: A
Why: Its kernel is only zero, so the map is injective, but the odd integers are never outputs, so it is not surjective onto the integers.
Elimination
Eliminate the wrong options
Which of these subgroups of the permutations of three symbols is normal?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The alternating subgroup has index two, and every index-two subgroup is normal, whereas each transposition subgroup is carried to a different one by conjugation.
Check
Work inside the symmetric group on three symbols and recall the conjugation test for normality.
Check your understanding
Which of these subgroups of the permutations of three symbols is normal?
Answer: A
Why: The alternating subgroup has index two, and every index-two subgroup is normal, whereas each transposition subgroup is carried to a different one by conjugation.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: use a homomorphism to identify a quotient · A homomorphism preserves the operation · The equation lives in two different groups · A homomorphism is a translation dictionary · A homomorphism sends the identity to the identity. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A homomorphism preserves the operation, and from that one equation everything else follows: it sends the identity to the identity and inverses to inverses.
Its image is a subgroup of the target, and its kernel is a normal subgroup of the source that measures exactly how far the map is from injective.
A quotient group exists precisely when you quotient by a normal subgroup, because normality is what makes coset multiplication well-defined.
\[ G/\ker\varphi \;\cong\; \operatorname{im}\varphi \]
The First Isomorphism Theorem ties it together: collapse the kernel and you recover a perfect copy of the image. Structure-preserving map, kernel, quotient, isomorphism onto the image is the pattern that reappears for rings, vector spaces, and beyond.
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