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The even permutations as a subgroup of half the size, a census of the permutations of four letters by cycle type, and the theorem that every group is a group of permutations.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to say why exactly half the permutations of a set are even, compute the order of an alternating group, count the permutations of each cycle type and check the total, explain what a subgroup of index two is and why the odd permutations are not one, and state and build the proof of Cayley's theorem.
Parity was a number attached to one permutation. Taken across a whole symmetric group, it splits it exactly in two, and one of the halves is a subgroup. This lesson looks at that half, counts the elements of each shape, and then proves the theorem that makes the whole unit general rather than special.
The alternating group $A_n$ is the subgroup of even permutations inside $S_n$, of order $n!/2$ for $n \ge 2$. The index of a subgroup is the number of cosets, so $A_n$ has index $2$. The cycle type of a permutation is its multiset of cycle lengths, fixed letters included. Cayley's theorem says every group is isomorphic to a subgroup of a symmetric group.
The alternating group. The even permutations form a subgroup. They are exactly half of $S_n$: fix any transposition $\tau$; the map $\sigma \mapsto \tau\sigma$ changes parity and is its own inverse, so it matches the even permutations with the odd ones one for one. Hence $|A_n| = n!/2$ for $n \ge 2$, and $A_n$ has index $2$.
$A_3$ is cyclic of order $3$. $A_4$ has order $12$ and is the first alternating group that is not abelian. $A_n$ for $n \ge 5$ is simple — it has no normal subgroup but the trivial one and itself — which is the fact behind the unsolvability of the general quintic, and is proved in Abstract algebra II rather than here.
Counting by cycle type. The permutations of $n$ letters split by cycle type, and each type is counted the same way: choose which letters each cycle uses, arrange them, and divide by the number of writings each cycle has. For $S_4$ the five types number $1, 6, 3, 8, 6$, adding to $24$.
Cayley's theorem. For $g \in G$ let $\lambda_g(x) = gx$. Cancellation makes $\lambda_g$ a bijection of the set $G$; associativity makes $\lambda_{gh} = \lambda_g \lambda_h$; and $\lambda_g = \mathrm{id}$ forces $g = e$. So $g \mapsto \lambda_g$ embeds $G$ in the permutations of its own elements. A group of order $n$ therefore sits inside $S_n$.
An algebraic fact is about the operation, not about what the elements happen to be. The same group turns up as rotations of a square, as residues under addition and as matrices, and a theorem proved once from the axioms holds in all of them at once.
Another way: picture
Lay the $n!$ permutations out in two columns, even on the left and odd on the right. Pick one transposition and draw a line from each permutation to what multiplying by it gives: every line crosses from one column to the other, and following two lines returns you to where you began. The columns are therefore the same height. That picture is the whole proof that exactly half of them are even.
Another way: steps
To count a family of permutations by cycle type: 1. Choose which letters the first cycle uses. 2. Count the distinct cycles on those letters: arrangements divided by the cycle length. 3. Repeat for the remaining cycles, dividing by the number of ways cycles of equal length can be swapped. 4. Add the counts over all types and check the total is $n!$.
The theorem is a statement of principle, not a computational tool. It says the symmetric groups are universal: anything true of every subgroup of every $S_n$ is true of every group. That is why the next three units may be read as being about permutations without loss, and why $S_n$ is worth this much attention.
What it is not good for is making groups smaller. The embedding it produces is wasteful: a group of order $n$ lands in $S_n$, which has $n!$ elements, so a group of order $8$ is placed inside a group of order $40320$. Better embeddings usually exist — $D_4$ sits comfortably inside $S_4$ — and finding the smallest is a real question that the theorem does not answer.
The construction repays a second look, because it reappears throughout the subject. The map $\lambda_g$ is the row of $g$ in the Cayley table, read as a rearrangement of the headings. So the theorem is the observation from lesson 2 — every row of a Cayley table is a permutation of the elements — taken seriously and turned into an isomorphism. In Abstract algebra II the same idea, with $G$ acting on something other than itself, becomes the theory of group actions, and Cayley's theorem is the case where the thing acted on is the group.
A subgroup of index $2$, like $A_n$, has one further property worth naming now: it is normal, because its two left cosets and its two right cosets are the same pair of sets — the subgroup, and everything else. Unit 3 makes that the definition that matters.
Saying the odd permutations form a subgroup too. They do not: the identity is even, so the odd permutations do not contain it, and a product of two odd permutations is even.
Forgetting to divide when counting cycles. There are $4!$ ways to arrange four letters in a bracket but only $4!/4 = 6$ distinct four-cycles, because each is written four times over.
*Reading Cayley's theorem as every group is a symmetric group. It says every group is isomorphic to a subgroup* of one, which is a much weaker and much more useful claim.
Expecting $A_n$ to be abelian because it is smaller. $A_4$ is not abelian, and $A_4$ also has no subgroup of order $6$ — so it is the standard counterexample to the converse of Lagrange, which arrives in the next unit.
Thinking half of $S_1$ is even. $|A_n| = n!/2$ needs $n \ge 2$: there is no transposition available on one letter, and $S_1$ is the trivial group.
The identity, $1$ of them; the double transpositions $(1\,2)(3\,4)$, $(1\,3)(2\,4)$, $(1\,4)(2\,3)$, $3$ of them.
Both types are even.
The three-cycles: $8$ of them, each even because a cycle of length $3$ takes two transpositions.
Even, despite being cycles of odd length.
Total $1 + 3 + 8 = 12 = 24/2$, as it must be. The transpositions and four-cycles, $6 + 6 = 12$, make up the odd half.
The two halves, counted.
Take $\mathbb{Z}_3 = \{0, 1, 2\}$ and label the elements $1, 2, 3$ in that order. Left multiplication by $1$ means adding $1$.
Left multiplication, additively.
Adding $1$ sends $0 \to 1 \to 2 \to 0$, which as a permutation of the labels is the three-cycle $(1\,2\,3)$.
One element, one permutation.
Adding $2$ gives $(1\,3\,2)$ and adding $0$ gives the identity, so $\mathbb{Z}_3$ is isomorphic to $\{e, (1\,2\,3), (1\,3\,2)\} = A_3$.
The image is a subgroup of $S_3$.
All five letters are used, so there is no choice of which letters: just arrangements of all five in a bracket, $5! = 120$ of them.
No letters to choose.
But each five-cycle is written five times over, once for each starting letter.
Divide by the length.
So there are $120/5 = 24$ of them. Each is even, since a cycle of length $5$ takes four transpositions, so all $24$ lie in $A_5$.
Fill in how many even permutations there are of $3$, $4$, $5$ and $6$ letters. The first column, already given, is the total number of permutations.
| Permutations altogether | Even permutations | |
|---|---|---|
| 3 letters | 6 | |
| 4 letters | 24 | |
| 5 letters | 120 | |
| 6 letters | 720 |
How many even permutations of $3$ letters are there?
Answer:
Build the proof of Cayley's theorem: every group is isomorphic to a group of permutations.
This task has no paper form; do it on a device.
Match each group to the number of elements it has.
| $4$ | $8$ | $12$ | $24$ | $6$ | |
|---|---|---|---|---|---|
| All the permutations of four letters | |||||
| The even permutations of four letters | |||||
| The symmetries of a square | |||||
| The integers modulo four under addition |
Select every statement about the even permutations of $n$ letters that is true, for $n$ at least $4$.
This task has no paper form; do it on a device.
The $24$ permutations of four letters fall into five cycle types. Say how many permutations there are of each.
| How many | |
|---|---|
| The identity | 1 |
| A single transposition | |
| Two disjoint transpositions | |
| A cycle of length three | |
| A cycle of length four |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Fill in how many even permutations there are of $3$, $4$, $5$ and $6$ letters. The first column, already given, is the total number of permutations.
| Permutations altogether | Even permutations | |
|---|---|---|
| 3 letters | 6 | |
| 4 letters | 24 | |
| 5 letters | 120 | |
| 6 letters | 720 |
You can count the even permutations, count a cycle type, and state Cayley's theorem. Say in your own words why exactly half of the permutations are even, and why the odd ones are not a subgroup. Next: cosets, which turn half of them into a general way of dividing a group up.
9. Your turn: how many permutations of five letters are cycles of length five?, step 3