This 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.
Subject: Foundations of Higher Mathematics · 118 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Decide whether a subset is a subgroup using the one-step test.
2. Build the cyclic subgroup generated by an element and read off the element's order.
3. List the left cosets of a subgroup and see them partition the group into equal-size tiles.
4. State and prove Lagrange's theorem, and use its corollaries to derive Fermat's little theorem.
Warm-up
Discussion prompt
Before we open Subgroups, Cosets & Lagrange's Theorem: without looking back, what was the main idea of Groups: Axioms & First Examples, 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 presents the four group axioms as the upgrade from a monoid, then proves that the identity and inverses are unique and covers the cancellation law and the socks-and-shoes rule. It distinguishes the order of a group from the order of an element, and abelian from non-abelian groups, and works the standard first examples: the additive integers, the integers mod n, the units mod n, the symmetric group, and the dihedral symmetries of a square. It targets the real traps: forgetting closure or inverses when calling a subset a group, assuming every group is abelian, confusing the order of a group with the order of an element, and thinking that the integers mod n form a group under multiplication.
Concept
A subgroup is a subset of a group that is itself a group under the same operation.
\[ H \leq G \iff H \subseteq G \text{ and } (H, \cdot) \text{ is a group} \]
subgroup — A subset H of a group G that contains the identity, is closed under the operation, and contains the inverse of each of its elements. Written H is less than or equal to G.
Counterexample
Discussion prompt
A subgroup is a subset of a group that is itself a group under the same operation.
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.
Intuition
You are not inventing a new operation. You take the group's operation and ask: does this subset stay closed and complete under it?
Think of a smaller machine built from the same gears: whatever combinations you perform inside it never spit out a part that lives outside it.
Every group has two subgroups for free: the whole group, and the one-element set holding just the identity.
Analogy
Discussion prompt
Explain Same operation, self-contained 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:
You are not inventing a new operation. You take the group's operation and ask: does this subset stay closed and complete under it?
Concept
Two subgroups always exist and are usually not what we care about.
\[ \{e\} \leq G \qquad \text{and} \qquad G \leq G \]
The one-element subgroup is called trivial; the whole group is the improper subgroup. The interesting subgroups sit strictly between these.
Explain it
Discussion prompt
Explain The trivial and improper subgroups 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 subgroups always exist and are usually not what we care about.
Concept
To verify a subset is a subgroup by hand, check three properties.
\[ \text{(1) identity: } e \in H \]
\[ \text{(2) closure: } a, b \in H \implies ab \in H \]
\[ \text{(3) inverses: } a \in H \implies a^{-1} \in H \]
Ranking
Put in order
Put the moves of Verify a subset is a subgroup 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 additive identity of the group is 0, and 0 is listed in H.
Worked example
Work in the group of integers modulo 6 under addition. Test the subset containing 0, 2, and 4.
\[ G = \mathbb{Z}/6\mathbb{Z}, \quad H = \{0, 2, 4\} \]
Confirm the identity is in H
Why: The additive identity of the group is 0, and 0 is listed in H.
Check closure on every sum
Why: Add elements of H mod 6 and confirm the result stays inside H.
\[ 2+4 = 0, \; 2+2 = 4, \; 4+4 = 2 \pmod 6 \]
Check each element has its inverse in H
Why: The additive inverse of x mod 6 is 6 minus x; each lands back in H.
\[ -0 = 0,\; -2 = 4,\; -4 = 2 \in H \]
Verify all three axioms hold, so H is a subgroup
Why: Identity present, closed under addition, inverses present. Hence H is a subgroup of order 3.
\[ H = \{0,2,4\} \leq \mathbb{Z}/6\mathbb{Z} \]
Picture it
Animation
Shows: Each line of the worked example "Verify a subset is a subgroup", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Add elements of H mod 6 and confirm the result stays inside H.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting claim: the natural numbers are a subgroup of the integers, because adding two of them gives another one.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The sum of two natural numbers is a natural number, and 0 is present.
Closure alone is not enough for an infinite subset. You must also check inverses.
Why: The sum of two natural numbers is a natural number, and 0 is present. Two of the three boxes are ticked.
Trap
Tempting claim: the natural numbers are a subgroup of the integers, because adding two of them gives another one.
\[ \mathbb{N} = \{0,1,2,\dots\} \subseteq \mathbb{Z} \]
Closure holds, so it looks done
Why: The sum of two natural numbers is a natural number, and 0 is present. Two of the three boxes are ticked.
\[ a, b \in \mathbb{N} \implies a + b \in \mathbb{N} \]
But inverses fail
Why: The additive inverse of 1 is negative 1, which is not a natural number. The set is not closed under inverses.
\[ 1 \in \mathbb{N}, \quad -1 \notin \mathbb{N} \]
Closure alone is not enough for an infinite subset. You must also check inverses.
The integers are the smallest subgroup here
Why: Once you include the inverse of every element, the closed set forced by 1 is all of the integers, not the naturals.
\[ \langle 1 \rangle = \mathbb{Z} \leq \mathbb{Z} \]
Note the finite rescue
Why: For a NONEMPTY FINITE subset, closure alone forces inverses (repeated addition cycles back). The trap only bites in the infinite case.
Notation
Annotate
From Trap: closed under the operation is not enough — read this one piece at a time. What is each part doing?
On: \( \mathbb{N} = \{0,1,2,\dots\} \subseteq \mathbb{Z} \)
Concept
Checking three axioms separately is wasteful. One condition bundles them together.
\[ H \leq G \iff H \neq \varnothing \text{ and } \forall a,b \in H,\; ab^{-1} \in H \]
one-step subgroup test — A nonempty subset H is a subgroup exactly when, for all a and b in H, the product of a with the inverse of b stays in H.
Fill the middle
Fill in the blanks
From Why one product does all the work — finish the line. Write what belongs on the right of the equals sign before you look.
aa^e \in H = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Taking the product of an element with its own inverse yields the identity, so the identity is in H.
Intuition
The single expression secretly checks identity, inverses, and closure in sequence.
Set a equal to b to extract the identity
Why: Taking the product of an element with its own inverse yields the identity, so the identity is in H.
\[ aa^{-1} = e \in H \]
Set a equal to the identity to extract inverses
Why: The product of e with the inverse of b is the inverse of b, so inverses are in H.
\[ eb^{-1} = b^{-1} \in H \]
Combine with inverses to recover closure
Why: Since the inverse of b is in H, replacing it gives the plain product, so H is closed.
\[ a(b^{-1})^{-1} = ab \in H \]
Step zero
Discussion prompt
One-step test on the even integers — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Confirm H is nonempty
Answer:
Worked example
Show the even integers form a subgroup of the integers under addition. In additive notation the test uses subtraction.
\[ H = 2\mathbb{Z} = \{\dots,-4,-2,0,2,4,\dots\} \]
Confirm H is nonempty
Why: Zero is even, so H contains at least one element.
Take two even numbers and form a minus b
Why: Write them as twice something; their difference is twice the difference of those somethings.
\[ a = 2m,\; b = 2n \implies a - b = 2(m-n) \]
Verify the difference is even, so the test passes
Why: The difference is 2 times an integer, hence even, hence in H. By the one-step test H is a subgroup.
\[ 2(m-n) \in 2\mathbb{Z} \implies 2\mathbb{Z} \leq \mathbb{Z} \]
Picture it
Animation
Shows: Each line of the worked example "One-step test on the even integers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The difference is 2 times an integer, hence even, hence in H. By the one-step test H is a subgroup.
Estimation
Predict first
Now test the odd integers under addition. We expect failure.
Commit before you compute: what does A subset that fails the test 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 S is not a subgroup
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. It misses the identity and is not closed under differences, so the odd integers are not a subgroup.
Worked example
Now test the odd integers under addition. We expect failure.
\[ S = \{\dots, -3, -1, 1, 3, \dots\} \]
Check the identity first
Why: The additive identity is 0, which is even, so 0 is not in S. Already the identity axiom fails.
\[ 0 \notin S \]
Confirm closure also breaks
Why: The one-step test needs a minus b in S; but the difference of two odd numbers is even.
\[ 3 - 1 = 2 \notin S \]
Verify S is not a subgroup
Why: It misses the identity and is not closed under differences, so the odd integers are not a subgroup.
Picture it
Animation
Shows: Each line of the worked example "A subset that fails the test", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The additive identity is 0, which is even, so 0 is not in S. Already the identity axiom fails.
Concept
Subgroups are not just examples; they combine. Overlapping two subgroups always yields a subgroup.
\[ H \leq G \text{ and } K \leq G \implies H \cap K \leq G \]
This is why there is always a smallest subgroup containing any given set: intersect all the subgroups that contain it.
Missing information
Discussion prompt
Use the one-step test on the intersection of two subgroups H and K of G.
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:
Both H and K contain the identity, so the identity lies in their intersection.
Worked example
Use the one-step test on the intersection of two subgroups H and K of G.
Show the intersection is nonempty
Why: Both H and K contain the identity, so the identity lies in their intersection.
\[ e \in H \text{ and } e \in K \implies e \in H \cap K \]
Take a and b in the intersection
Why: Membership in the intersection means membership in each subgroup separately.
\[ a, b \in H \cap K \implies a,b \in H \text{ and } a,b \in K \]
Apply the one-step test inside each subgroup
Why: Since H and K are subgroups, the product of a with the inverse of b lies in each of them.
\[ ab^{-1} \in H \text{ and } ab^{-1} \in K \]
Verify the product stays in the intersection
Why: Being in both H and K means being in the intersection, so the one-step test passes and the intersection is a subgroup.
\[ ab^{-1} \in H \cap K \implies H \cap K \leq G \]
Picture it
Animation
Shows: Each line of the worked example "Prove the intersection is a subgroup", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Being in both H and K means being in the intersection, so the one-step test passes and the intersection is a subgroup.
Concept
Pick a single element and take all of its powers, positive, negative, and zero. The result is always a subgroup.
\[ \langle g \rangle = \{ g^k : k \in \mathbb{Z} \} \]
cyclic subgroup — The set of all integer powers of a fixed element g. It is the smallest subgroup containing g, and it is generated by that one element.
Definition probe
Sort into buckets
Every line below is part of the definition of subgroup or of cyclic subgroup — one or the other, never both. Put each where it belongs.
Intuition
Start at the identity and keep applying the same element. You trace out an orbit that eventually closes into a loop (finite case) or marches off forever (infinite case).
In a finite group the loop must close: there are only finitely many places to land, so a power must repeat, and the first repeat brings you back to the identity.
Fill the middle
Fill in the blanks
From Build a cyclic subgroup — finish the line. Write what belongs on the right of the equals sign before you look.
\langle 2 \rangle = \{0, 2, 4\}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Add 2 to itself repeatedly and reduce mod 6 until you return to the start.
Worked example
In the integers modulo 6 under addition, powers mean repeated addition. Generate the subgroup from the element 2.
List the multiples of 2 modulo 6
Why: Add 2 to itself repeatedly and reduce mod 6 until you return to the start.
\[ 2,\; 2+2=4,\; 4+2=0 \]
Stop when you hit the identity
Why: Reaching 0 closes the loop; further additions just repeat the cycle.
\[ \langle 2 \rangle = \{0, 2, 4\} \]
Verify the generated set matches the earlier subgroup
Why: The cyclic subgroup generated by 2 is exactly the subgroup we tested by hand, confirming it is a subgroup of order 3.
Picture it
Animation
Shows: Each line of the worked example "Build a cyclic subgroup", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The cyclic subgroup generated by 2 is exactly the subgroup we tested by hand, confirming it is a subgroup of order 3.
Concept
The order of an element is how many steps it takes to first return to the identity.
\[ \operatorname{ord}(g) = \min \{ n \geq 1 : g^n = e \} \]
If no such positive power exists, the element has infinite order.
Intuition
The order counts the distinct stops on the orbit before it loops back. It is literally the length of the cycle you traced when generating the subgroup.
So the order of an element and the size of the subgroup it generates are the same number, said two different ways.
Concept
The two viewpoints coincide exactly.
\[ \operatorname{ord}(g) = |\langle g \rangle| \]
This is the bridge we will need for Lagrange: once we know subgroup sizes divide the group's size, element orders will inherit the same divisibility.
Step zero
Discussion prompt
Orders of every element modulo 6 — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Track how many additions return each element to 0
Answer:
Worked example
Compute the additive order of each element of the integers modulo 6.
Track how many additions return each element to 0
Why: Add the element to itself, counting steps until reaching the identity 0.
| element | cycle | order |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 1,2,3,4,5,0 | 6 |
| 2 | 2,4,0 | 3 |
| 3 | 3,0 | 2 |
| 4 | 4,2,0 | 3 |
| 5 | 5,4,3,2,1,0 | 6 |
Read off the orders
Why: The step counts give the orders 1, 6, 3, 2, 3, 6 for the elements 0 through 5.
Verify every order divides 6
Why: The orders 1, 2, 3, 6 are exactly the divisors of 6. This previews Lagrange's corollary that element order divides group order.
\[ 1,2,3,6 \mid 6 \]
Picture it
Animation
Shows: Each line of the worked example "Orders of every element modulo 6", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The orders 1, 2, 3, 6 are exactly the divisors of 6. This previews Lagrange's corollary that element order divides group order.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting claim: the order of an element equals the order of the group it lives in.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: A student sees the group has 6 elements and assumes every element also has order 6.
The element order is a property of the element; the group order is the count of all elements. They need not match.
Why: A student sees the group has 6 elements and assumes every element also has order 6.
Trap
Tempting claim: the order of an element equals the order of the group it lives in.
Assume order of element equals 6 for all of them
Why: A student sees the group has 6 elements and assumes every element also has order 6.
\[ \operatorname{ord}(3) \overset{?}{=} |\mathbb{Z}/6\mathbb{Z}| = 6 \]
But 3 returns to 0 in only two steps
Why: Adding 3 twice gives 6, which is 0 mod 6, so the order of 3 is 2, not 6.
\[ 3 + 3 = 0 \implies \operatorname{ord}(3) = 2 \]
The element order is a property of the element; the group order is the count of all elements. They need not match.
Keep the two numbers distinct
Why: The group order is 6; the element orders range over 1, 2, 3, 6. Lagrange only forces the element order to DIVIDE the group order.
\[ \operatorname{ord}(g) \mid |G| \]
Translation
\( \operatorname{ord}(3) \overset{?}{=} |\mathbb{Z}/6\mathbb{Z}| = 6 \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Fix a subgroup H and an element g. The left coset is what you get by combining g with every element of H.
\[ gH = \{ gh : h \in H \} \]
left coset — The set g times H, formed by multiplying a fixed g on the left into every element of the subgroup H. In additive notation it is g plus H.
Intuition
Picture the subgroup as a tile sitting at the identity. Multiplying by g slides that whole tile to a new location without changing its shape or size.
Different shifts either land on exactly the same tile or on a completely disjoint one. There is never a partial overlap. That is the seed of the partition.
Picture it
Figure (svg): Two boxes: the left box labeled 0 plus H holds 0, 2, 4; the right box labeled 1 plus H holds 1, 3, 5, together covering all six residues.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Take the subgroup of order 3 inside the integers modulo 6, and slide it by each group element.
Worked example
Take the subgroup of order 3 inside the integers modulo 6, and slide it by each group element.
\[ H = \{0,2,4\} \leq \mathbb{Z}/6\mathbb{Z} \]
Shift H by 0
Why: Adding 0 to every element of H returns H itself.
\[ 0 + H = \{0,2,4\} \]
Shift H by 1
Why: Adding 1 to each element of H mod 6 gives the odd residues.
\[ 1 + H = \{1,3,5\} \]
Verify these two cosets tile the whole group
Why: The evens and the odds are disjoint and together give all six residues, so the cosets partition the group into two tiles of size 3.
Figure (svg): Two boxes: the left box labeled 0 plus H holds 0, 2, 4; the right box labeled 1 plus H holds 1, 3, 5, together covering all six residues.
Reverse engineer
Discussion prompt
Work backwards. The example finished here:
Verify these two cosets tile the whole group
What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.
Hint: Every quantity in the result had to enter somewhere. Account for each one.
Answer:
Take the subgroup of order 3 inside the integers modulo 6, and slide it by each group element.
Picture it
Figure (svg): A long bar split into four equal boxes, each a coset of equal size, tiling the whole group with no gaps or overlaps.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The picture generalizes: the left cosets of any subgroup carve the group into disjoint, equal-size pieces that cover everything.
Concept
The picture generalizes: the left cosets of any subgroup carve the group into disjoint, equal-size pieces that cover everything.
Figure (svg): A long bar split into four equal boxes, each a coset of equal size, tiling the whole group with no gaps or overlaps.
Two cosets are either identical or disjoint. This dichotomy is what makes them the blocks of a genuine partition.
Concept
There is a clean membership test for coset equality that avoids listing elements.
\[ gH = g'H \iff g^{-1}g' \in H \]
In additive notation this reads: two cosets coincide exactly when the difference of their representatives lies in the subgroup.
Hypothesis
Predict first
Decide whether two cosets are equal 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: Form the difference of the two representatives
Why: The additive test asks whether 3 minus 1 lands in H.
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
Using the subgroup of order 3 in the integers modulo 6, compare the cosets built from 1 and from 3.
\[ H = \{0,2,4\}, \quad 1+H \;\text{vs}\; 3+H \]
Form the difference of the two representatives
Why: The additive test asks whether 3 minus 1 lands in H.
\[ 3 - 1 = 2 \in H \]
Conclude the cosets are equal
Why: Since the difference 2 lies in the subgroup, the two cosets are the same set.
\[ 3 + H = 1 + H = \{1,3,5\} \]
Verify by direct listing
Why: Adding 3 to each of 0, 2, 4 gives 3, 5, 1, the same set as 1 plus H. The test agrees with the explicit computation.
\[ 3+H = \{3,5,1\} = \{1,3,5\} \]
Picture it
Animation
Shows: Each line of the worked example "Decide whether two cosets are equal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Adding 3 to each of 0, 2, 4 gives 3, 5, 1, the same set as 1 plus H. The test agrees with the explicit computation.
Concept
Any element of a coset can name it. The chosen name is a representative.
A set with exactly one representative from each coset is a transversal. It is a menu of tile labels, one per tile.
Choosing a different representative never changes the coset, only its label. That freedom is exactly the coset-equality test in disguise.
Ranking
Put in order
Put the moves of Cosets in the integers 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. Adding a fixed number to every multiple of 3 gives a shifted arithmetic progression.
Worked example
Take the multiples of 3 as a subgroup of the integers under addition, and slide it by 0, 1, and 2.
\[ H = 3\mathbb{Z} \leq \mathbb{Z} \]
Slide by 0 and by 1
Why: Adding a fixed number to every multiple of 3 gives a shifted arithmetic progression.
\[ 0 + 3\mathbb{Z} = \{\dots,-3,0,3,\dots\}, \; 1 + 3\mathbb{Z} = \{\dots,-2,1,4,\dots\} \]
Slide by 2 and stop
Why: Sliding by 3 would land back on the subgroup, so only three distinct cosets appear.
\[ 2 + 3\mathbb{Z} = \{\dots,-1,2,5,\dots\} \]
Verify the cosets are the residue classes mod 3
Why: The three cosets are exactly the numbers leaving remainder 0, 1, or 2 on division by 3; they partition all integers.
\[ \mathbb{Z} = 3\mathbb{Z} \;\sqcup\; (1+3\mathbb{Z}) \;\sqcup\; (2+3\mathbb{Z}) \]
Picture it
Animation
Shows: Each line of the worked example "Cosets in the integers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The three cosets are exactly the numbers leaving remainder 0, 1, or 2 on division by 3; they partition all integers.
Concept
Define two elements to be related when one is the other slid by an element of the subgroup. This relation is reflexive, symmetric, and transitive.
\[ a \sim b \iff a^{-1}b \in H \]
Its equivalence classes are precisely the left cosets. That single fact is what turns cosets into a partition.
Fill the middle
Fill in the blanks
From Prove the cosets partition the group — finish the line. Write what belongs on the right of the equals sign before you look.
a^(a^{-1}b)(b^{-1}c) \in Hb,\, b^___c \in H \implies a^___c = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The identity is in H, so a is related to itself.
Worked example
Show the relation of sharing a coset is an equivalence relation, so its classes tile the group.
Reflexive
Why: The identity is in H, so a is related to itself.
\[ a^{-1}a = e \in H \implies a \sim a \]
Symmetric
Why: If the connecting element is in H, so is its inverse, since H is a subgroup.
\[ a^{-1}b \in H \implies (a^{-1}b)^{-1} = b^{-1}a \in H \]
Transitive
Why: The product of two elements of H stays in H by closure.
\[ a^{-1}b,\, b^{-1}c \in H \implies a^{-1}c = (a^{-1}b)(b^{-1}c) \in H \]
Verify the classes are exactly the left cosets
Why: The class of a is the set of all ah for h in H, which is the coset aH. Equivalence classes always partition, so the cosets partition G.
\[ [a] = aH, \quad G = \bigsqcup_i g_i H \]
Picture it
Animation
Shows: Each line of the worked example "Prove the cosets partition the group", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The class of a is the set of all ah for h in H, which is the coset aH. Equivalence classes always partition, so the cosets partition G.
Concept
The tiles are not just disjoint; they are all the same size, equal to the size of the subgroup.
\[ |gH| = |H| \quad \text{for every } g \in G \]
The reason is a bijection between the subgroup and each coset, built by left multiplication.
Estimation
Predict first
Fix g and define the map that sends an element of the subgroup to its shift. Show it is a bijection from H onto the coset gH.
Commit before you compute: what does The shift map is a bijection 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 size equality follows
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. A bijection between H and gH forces them to have the same number of elements, so every coset has size equal to the size of H.
Worked example
Fix g and define the map that sends an element of the subgroup to its shift. Show it is a bijection from H onto the coset gH.
\[ \varphi : H \to gH, \quad \varphi(h) = gh \]
Show it is onto
Why: Every element of gH has the form gh by definition, so it is hit by h.
\[ gh \in gH \text{ is the image of } h \]
Show it is one-to-one using cancellation
Why: If two shifts agree, left-cancel g to conclude the inputs agree.
\[ gh_1 = gh_2 \implies h_1 = h_2 \]
Verify the size equality follows
Why: A bijection between H and gH forces them to have the same number of elements, so every coset has size equal to the size of H.
\[ |gH| = |H| \]
Concept
Sliding on the other side gives right cosets. The number of cosets, left or right, is the same, and it has a name.
\[ Hg = \{ hg : h \in H \}, \qquad [G:H] = \text{number of cosets} \]
index — The index of H in G, written the bracket G colon H, is the number of distinct left cosets of H in G. It equals the number of right cosets too.
Intuition
You have a floor (the group) cut into identical tiles (the cosets). Count the tiles and multiply by the size of one tile to recover the total area.
Figure (svg): Four equal tiles each holding three dots; the caption shows four tiles times three dots equals twelve, illustrating the group size as index times subgroup size.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting claim: the left coset and the right coset of an element are always the same set.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Multiply the 3-cycle on the left into each element of H.
The two sets share the 3-cycle but differ in their other element, so they are not equal.
Why: Multiply the 3-cycle on the left into each element of H.
Trap
Tempting claim: the left coset and the right coset of an element are always the same set.
Work in the symmetric group on three symbols with the two-element subgroup generated by the swap of 1 and 2.
\[ G = S_3, \quad H = \{ e, (1\,2) \} \]
Compute the left coset of the 3-cycle
Why: Multiply the 3-cycle on the left into each element of H.
\[ (1\,2\,3)H = \{ (1\,2\,3),\, (1\,3) \} \]
Compute the right coset
Why: Multiply the same 3-cycle on the right; the second element changes.
\[ H(1\,2\,3) = \{ (1\,2\,3),\, (2\,3) \} \]
The two sets share the 3-cycle but differ in their other element, so they are not equal.
\[ (1\,2\,3)H \neq H(1\,2\,3) \]
Left equals right exactly for normal subgroups
Why: The coincidence of left and right cosets for every g is the definition of a normal subgroup; H here is not normal, so they differ. Counting still works either way.
Notation
Annotate
From Trap: left and right cosets need not agree — read this one piece at a time. What is each part doing?
On: \( (1\,2\,3)H \neq H(1\,2\,3) \)
Concept
For a finite group, the size of any subgroup divides the size of the whole group.
\[ H \leq G,\; |G| < \infty \implies |H| \text{ divides } |G| \]
\[ |G| = [G:H]\,|H| \]
Intuition
Two facts do all the work: the cosets are disjoint and cover the group, and each has the same size as the subgroup.
So the total count is the number of cosets multiplied by the shared tile size. Divisibility is not a coincidence; it is arithmetic forced by the partition.
Step zero
Discussion prompt
Assemble the proof of Lagrange — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Partition the group into its left cosets
Answer:
Worked example
Combine the partition and the equal-size facts into a one-line count.
Partition the group into its left cosets
Why: The same-coset equivalence relation splits G into disjoint cosets that cover it.
\[ G = g_1H \sqcup g_2 H \sqcup \cdots \sqcup g_k H, \quad k = [G:H] \]
Use that each coset has size equal to the subgroup
Why: The shift map is a bijection, so every coset contributes exactly the size of H.
\[ |g_iH| = |H| \text{ for each } i \]
Add the sizes of the disjoint pieces
Why: Disjoint pieces let you sum their sizes to get the total.
\[ |G| = \sum_{i=1}^{k} |g_iH| = k\,|H| \]
Verify the divisibility conclusion
Why: Since the group size is the index times the subgroup size, the subgroup size divides the group size. This completes Lagrange.
\[ |G| = [G:H]\,|H| \implies |H| \text{ divides } |G| \]
Picture it
Animation
Shows: Each line of the worked example "Assemble the proof of Lagrange", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Since the group size is the index times the subgroup size, the subgroup size divides the group size. This completes Lagrange.
Concept
Because an element's order is the size of the subgroup it generates, Lagrange immediately controls it.
\[ \operatorname{ord}(g) = |\langle g \rangle| \text{ divides } |G| \]
This is why the orders we computed modulo 6 were forced to be among the divisors of 6.
Pattern
Predict first
The table runs: 1 | 12 | yes · 2 | 6 | yes · 3 | 4 | yes · 4 | 3 | yes · 6 | 2 | yes
In Element orders divide the group order, given the rows so far: what is the next one — the row where element is 0?
Correct: 0 | 1 | yes
| element | order | divides 12? |
|---|---|---|
| 1 | 12 | yes |
| 2 | 6 | yes |
| 3 | 4 | yes |
| 4 | 3 | yes |
| 6 | 2 | yes |
| 0 | 1 | yes |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Count additions to reach 0 for each element; the order equals 12 divided by the greatest common divisor of the element with 12.
Worked example
In the integers modulo 12 under addition, compute several element orders and compare with the divisors of 12.
Tabulate a few orders
Why: Count additions to reach 0 for each element; the order equals 12 divided by the greatest common divisor of the element with 12.
| element | order | divides 12? |
|---|---|---|
| 1 | 12 | yes |
| 2 | 6 | yes |
| 3 | 4 | yes |
| 4 | 3 | yes |
| 6 | 2 | yes |
| 0 | 1 | yes |
Compare against the divisors of 12
Why: The divisors of 12 are 1, 2, 3, 4, 6, and 12, exactly the orders that appear.
\[ \{1,2,3,4,6,12\} = \text{divisors of } 12 \]
Verify no order 5 or 7 ever appears
Why: Since 5 and 7 do not divide 12, Lagrange forbids any element of those orders, matching the table.
Picture it
Animation
Shows: Each line of the worked example "Element orders divide the group order", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Since 5 and 7 do not divide 12, Lagrange forbids any element of those orders, matching the table.
Concept
Every element, raised to the power equal to the group's size, collapses to the identity.
\[ g^{|G|} = e \quad \text{for all } g \in G \]
The reason: the order of g divides the group size, so the group size is a whole number of full loops back to the identity.
Intuition
The corollaries all run through the same pipe. Turn an element into the subgroup it generates, apply Lagrange to that subgroup, and translate the divisibility back into a statement about the element.
Element order, the identity power, prime-order structure, and Fermat's theorem are all this one move applied in different costumes.
Concept
If a group has a prime number of elements, it must be generated by a single element.
\[ |G| = p \text{ prime} \implies G \text{ is cyclic} \]
A prime has almost no divisors, which leaves a nontrivial subgroup no room except to be the whole group.
Missing information
Discussion prompt
Let the group have p elements, with p prime, and pick any element other than the identity.
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:
Its size divides p by Lagrange, so it is either 1 or p.
Worked example
Let the group have p elements, with p prime, and pick any element other than the identity.
\[ |G| = p, \quad g \neq e \]
Look at the subgroup generated by g
Why: Its size divides p by Lagrange, so it is either 1 or p.
\[ |\langle g \rangle| \text{ divides } p \implies |\langle g \rangle| \in \{1, p\} \]
Rule out size one
Why: The subgroup contains g, which is not the identity, so it has more than one element.
\[ g \in \langle g \rangle, \; g \neq e \implies |\langle g \rangle| > 1 \]
Verify the subgroup is everything
Why: The only remaining option is size p, so the subgroup generated by g is all of G, meaning G is cyclic with generator g.
\[ \langle g \rangle = G \]
Picture it
Animation
Shows: Each line of the worked example "Prove a prime-order group is cyclic", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The only remaining option is size p, so the subgroup generated by g is all of G, meaning G is cyclic with generator g.
Concept
Apply the identity-power corollary to the group of nonzero residues modulo a prime, whose size is one less than the prime.
\[ |(\mathbb{Z}/p\mathbb{Z})^{\times}| = p - 1 \]
\[ \gcd(a,p)=1 \implies a^{\,p-1} \equiv 1 \pmod p \]
Explain it
Discussion prompt
Explain Fermat's little theorem from Lagrange 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:
Apply the identity-power corollary to the group of nonzero residues modulo a prime, whose size is one less than the prime.
Estimation
Predict first
Use Fermat's little theorem to evaluate a large power of 3 modulo the prime 7.
Commit before you compute: what does Derive and apply Fermat 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 by re-reducing
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. Check that the sixth power really is 1: seven hundred twenty-nine is one more than a multiple of seven.
Worked example
Use Fermat's little theorem to evaluate a large power of 3 modulo the prime 7.
\[ \text{Compute } 3^{100} \bmod 7 \]
Invoke Fermat to collapse the exponent's period
Why: Since 7 is prime and 3 is not a multiple of it, the sixth power is the identity, so exponents matter only modulo 6.
\[ 3^{6} \equiv 1 \pmod 7 \]
Reduce the exponent modulo 6
Why: One hundred is sixteen sixes plus four, so the effective exponent is four.
\[ 100 = 6\cdot 16 + 4 \implies 3^{100} \equiv 3^{4} \pmod 7 \]
Compute the small power
Why: Three to the fourth is eighty-one, which leaves remainder four on division by seven.
\[ 3^{4} = 81 \equiv 4 \pmod 7 \]
Verify by re-reducing
Why: Check that the sixth power really is 1: seven hundred twenty-nine is one more than a multiple of seven. Final answer four.
\[ 3^{6} = 729 = 7\cdot 104 + 1 \equiv 1 \pmod 7 \]
Picture it
Animation
Shows: Each line of the worked example "Derive and apply Fermat", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that the sixth power really is 1: seven hundred twenty-nine is one more than a multiple of seven. Final answer four.
Concept
When the modulus is not prime, replace the prime-minus-one count with the number of residues coprime to the modulus, the totient.
\[ \gcd(a,n)=1 \implies a^{\varphi(n)} \equiv 1 \pmod n \]
It is the identity-power corollary applied to the group of units modulo the modulus. Fermat is the special case when the modulus is prime.
Analogy
Discussion prompt
Explain Euler's theorem: the same idea for any modulus 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:
When the modulus is not prime, replace the prime-minus-one count with the number of residues coprime to the modulus, the totient.
Step zero
Discussion prompt
Order of an element in the units mod 7 — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: List the powers of 2 modulo 7
Answer:
Worked example
Find the order of 2 in the multiplicative group of nonzero residues modulo 7, and confirm it obeys Lagrange.
\[ G = (\mathbb{Z}/7\mathbb{Z})^{\times}, \quad |G| = 6 \]
List the powers of 2 modulo 7
Why: Multiply by 2 and reduce until reaching 1.
\[ 2^1 = 2,\; 2^2 = 4,\; 2^3 = 8 \equiv 1 \pmod 7 \]
Read off the order
Why: The first power equal to 1 is the third, so the order of 2 is 3.
\[ \operatorname{ord}(2) = 3 \]
Verify the order divides the group size and Fermat holds
Why: Three divides six as Lagrange demands, and the sixth power is the cube squared, which is one. Both checks pass.
\[ 3 \mid 6, \quad 2^6 = (2^3)^2 \equiv 1 \pmod 7 \]
Picture it
Animation
Shows: Each line of the worked example "Order of an element in the units mod 7", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Three divides six as Lagrange demands, and the sixth power is the cube squared, which is one. Both checks pass.
Ranking
Put in order
Put the moves of Compute an index 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. Add 4 repeatedly modulo 12 until returning to 0.
Worked example
Find the index of the subgroup generated by 4 inside the integers modulo 12 under addition.
\[ G = \mathbb{Z}/12\mathbb{Z}, \quad H = \langle 4 \rangle \]
Generate the subgroup
Why: Add 4 repeatedly modulo 12 until returning to 0.
\[ \langle 4 \rangle = \{0, 4, 8\}, \quad |H| = 3 \]
Divide the group size by the subgroup size
Why: Lagrange gives the index as the group order divided by the subgroup order.
\[ [G:H] = \frac{|G|}{|H|} = \frac{12}{3} = 4 \]
Verify by listing the four cosets
Why: The four cosets partition the twelve residues into four tiles of three, confirming the index is 4.
\[ \{0,4,8\},\{1,5,9\},\{2,6,10\},\{3,7,11\} \]
Picture it
Animation
Shows: Each line of the worked example "Compute an index", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The four cosets partition the twelve residues into four tiles of three, confirming the index is 4.
Concept
Every group, no matter how abstract, is secretly a group of permutations. Left multiplication by each element shuffles the group.
\[ G \hookrightarrow S_{|G|} \]
So the symmetric groups are universal: understanding permutation groups is, in principle, understanding all finite groups.
Counterexample
Discussion prompt
Every group, no matter how abstract, is secretly a group of permutations. Left multiplication by each element shuffles the group.
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.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting claim: if a number divides the group's size, there must be a subgroup of exactly that size.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It has twelve elements, and six divides twelve, so the claim predicts a subgroup of order six.
Lagrange runs one direction only: subgroup sizes divide the group size, not the reverse.
Why: It has twelve elements, and six divides twelve, so the claim predicts a subgroup of order six.
Trap
Tempting claim: if a number divides the group's size, there must be a subgroup of exactly that size.
\[ d \mid |G| \overset{?}{\implies} \exists\, H \leq G,\; |H| = d \]
Take the alternating group on four symbols
Why: It has twelve elements, and six divides twelve, so the claim predicts a subgroup of order six.
\[ |A_4| = 12, \quad 6 \mid 12 \]
But no such subgroup exists
Why: A careful check of its elements shows the alternating group on four symbols has no subgroup of order six at all.
Lagrange runs one direction only: subgroup sizes divide the group size, not the reverse.
Keep the implication one-way
Why: Divisibility is necessary for a subgroup to exist at that size, but not sufficient. Partial converses hold only under extra hypotheses, such as Cauchy's theorem for prime divisors or the Sylow theorems for prime powers.
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.
Pattern
1. To test a subgroup, use the one-step test
Why: Check the subset is nonempty and that the product of any element with the inverse of another stays inside.
2. To find an element's order, generate its cyclic subgroup
Why: The order is the length of the loop back to the identity, which equals the size of the generated subgroup.
3. To count cosets, divide the group size by the subgroup size
Why: The index is the group order over the subgroup order, since cosets are equal-size tiles that partition the group.
4. To bound orders, invoke Lagrange
Why: Any subgroup or element order must divide the group order; use this to rule out impossible sizes.
5. To evaluate large powers modulo a prime, use Fermat
Why: Exponents matter only modulo the group size, so reduce the exponent before computing.
Real world
Discussion prompt
Outside this lesson: where does Subgroups, Cosets & Lagrange's Theorem 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 The subgroup, coset, and Lagrange toolkit 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 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.
Elimination
Eliminate the wrong options
Which of these subsets is a subgroup of the integers under addition?
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 even integers contain 0, are closed under addition, and contain the negative of each even number, so they pass the one-step test and form a subgroup.
Check
Consider subsets of the integers under addition and decide which one is a subgroup.
Check your understanding
Which of these subsets is a subgroup of the integers under addition?
Answer: A
Why: The even integers contain 0, are closed under addition, and contain the negative of each even number, so they pass the one-step test and form a subgroup.
Check
Work in the integers modulo 6 with the subgroup consisting of 0, 2, and 4.
Check your understanding
How many distinct left cosets does the subgroup of size 3 have in the integers modulo 6?
Answer: A
Why: The index is the group size divided by the subgroup size, which is 6 divided by 3, giving 2 cosets: the evens and the odds.
Check
A group has exactly 15 elements. Use Lagrange to rule out an impossible element order.
Check your understanding
In a group of order 15, which of these cannot be the order of an element?
Answer: A
Why: By Lagrange the order of an element must divide the group order 15, whose divisors are 1, 3, 5, and 15; since 6 is not among them, no element can have order 6.
Check
Use Fermat's little theorem with the prime 13 to evaluate a power of 2.
Check your understanding
What is 2 raised to the 12th power, taken modulo 13?
Answer: A
Why: Since 13 is prime and 2 is not a multiple of it, Fermat's little theorem says 2 raised to the power 13 minus 1, that is the 12th power, is congruent to 1 modulo 13.
Prediction
Predict first
For a group of order 12, which statement is guaranteed true?
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: The order of every element divides 12
Why: Lagrange forces every element order and subgroup order to divide the group order 12, so option A always holds; the others can fail.
Check
A finite group has order 12. Decide which conclusion is actually forced.
Check your understanding
For a group of order 12, which statement is guaranteed true?
Answer: A
Why: Lagrange forces every element order and subgroup order to divide the group order 12, so option A always holds; the others can fail.
Prediction
Predict first
Which coset equals the one obtained by sliding the subgroup of size 3 by the element 5?
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: The coset built from 1
Why: Sliding by 5 gives the set 5, 1, 3, which is the odd residues, the same coset as the one built from 1; the equality test confirms 5 minus 1 equals 4, which is in the subgroup.
Check
In the integers modulo 6 with the subgroup of 0, 2, and 4, identify the coset built from 5.
Check your understanding
Which coset equals the one obtained by sliding the subgroup of size 3 by the element 5?
Answer: A
Why: Sliding by 5 gives the set 5, 1, 3, which is the odd residues, the same coset as the one built from 1; the equality test confirms 5 minus 1 equals 4, which is in the subgroup.
Elimination
Eliminate the wrong options
What is the order of the element 8 in the integers modulo 12 under addition?
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: Adding 8 repeatedly gives 8, then 4, then 0, so the cycle closes after three steps and the order is 3, which correctly divides 12.
Check
Work in the integers modulo 12 under addition and find the additive order of the element 8.
Check your understanding
What is the order of the element 8 in the integers modulo 12 under addition?
Answer: A
Why: Adding 8 repeatedly gives 8, then 4, then 0, so the cycle closes after three steps and the order is 3, which correctly divides 12.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The subgroup, coset, and Lagrange toolkit · A subgroup is a group living inside a group · Same operation, self-contained · The trivial and improper subgroups · The three things a subgroup must satisfy. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A subgroup is a subset that survives the one-step test: nonempty, and closed under the product of an element with another's inverse.
The cyclic subgroup generated by an element captures its order, which is the length of the loop back to the identity.
Left cosets are equal-size, disjoint tiles that partition the group, and their count is the index.
Lagrange's theorem says a subgroup's size divides the group's size, forcing element orders to divide the group order and yielding Fermat's little theorem as a corollary.
| Fact | What it gives you |
|---|---|
| One-step test | a fast subgroup check |
| Cosets partition G | equal tiles covering the group |
| Lagrange | subgroup size divides group size |
| Corollaries | element order divides order; Fermat |
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