Permutation groups: cycles, sign, conjugacy in S_n and A_n

In one sentence

Cycle type is the complete conjugacy invariant in $S_n$, so counting classes is counting partitions — and the two facts that catch people are that a cycle of even length is an odd permutation, and that classes can split on passing to $A_n$.

Why the exam asks it

Because $S_n$ is the group everything can be computed in, so questions are concrete: orders, signs, class sizes, centralisers. And because the sign of a cycle is off by one from the naive guess in a way that is easy to test.

The idea

Every permutation factors uniquely into disjoint cycles, and disjoint cycles commute. From that decomposition:

• Order $=$ lcm of the cycle lengths. So $(1\,2)(3\,4\,5)$ has order $6$.
• Sign $= (-1)^{n - c}$ where $c$ is the number of cycles counting fixed points. Equivalently a $k$-cycle is a product of $k-1$ transpositions, so its sign is $(-1)^{k-1}$.

That last point is the trap: a $k$-cycle is even exactly when $k$ is odd. A $3$-cycle is even, a $4$-cycle is odd. The parity of the cycle and the parity of the permutation are opposite.

$A_n$ is the kernel of the sign homomorphism, of index $2$ and order $n!/2$.

Conjugacy in $S_{n}$

Two permutations are conjugate in $S_n$ exactly when they have the same cycle type.

Conjugation relabels: $\sigma(a_1\,a_2\,\cdots)\sigma^{-1} = (\sigma(a_1)\,\sigma(a_2)\,\cdots)$, which is the most useful formula in the topic and worth being able to apply instantly.

So the conjugacy classes of $S_n$ correspond to partitions of $n$, and counting them is counting partitions: $5$ classes in $S_4$, $7$ in $S_5$.

The class of type $1^{a_1}2^{a_2}\cdots$ has size

$$\frac{n!}{\prod_k k^{a_k} a_k!},$$

which is the standard counting formula and appears directly in questions about class sizes and centraliser orders — the centraliser of an element has order $n!$ divided by its class size.

Classes can split in $A_{n}$

A class of $S_n$ lying in $A_n$ either stays one class or splits into two of equal size. It splits exactly when the element commutes with no odd permutation — equivalently when its cycle type has all parts odd and all distinct.

The standard case: the $5$-cycles in $S_5$ form one class of $24$; in $A_5$ they split into two classes of $12$. That splitting is what makes the class equation of $A_5$ read $60 = 1 + 15 + 20 + 12 + 12$, and it is why $A_5$ is simple — no sub-sum containing $1$ divides $60$ except $1$ and $60$ itself.

In $A_4$, the $3$-cycles split into two classes of $4$.

Generation

• $S_n$ is generated by transpositions; also by $(1\,2)$ and $(1\,2\,\cdots\,n)$.
• $A_n$ is generated by $3$-cycles.
• $Z(S_n) = \{e\}$ for $n \ge 3$.

Where intuition breaks

"A $4$-cycle is even because $4$ is even." A $k$-cycle has sign $(-1)^{k-1}$, so a $4$-cycle is odd. The most reliable sign error.

"Same order means conjugate." In $S_4$, $(1\,2)(3\,4)$ and $(1\,2)$ both have order $2$ and different cycle types, so they are not conjugate. Cycle type is the invariant, and order is only a consequence of it.

"Conjugate in $S_n$, so conjugate in $A_n$." The $5$-cycles in $A_5$. This is the topic's second big trap and it is what makes $A_5$'s class equation work.

"$A_n$ is generated by transpositions." Transpositions are odd, so they are not in $A_n$ at all. $3$-cycles generate it.

"$(1\,2\,3)$ and $(1\,3\,2)$ are different cycle types." Both are $3$-cycles, hence conjugate. They are inverses of each other, which is not a different type.

"The number of $k$-cycles in $S_n$ is $\binom{n}{k}$." It is $\dfrac{n!}{k\,(n-k)!}$ — choose the $k$ points and then count the $(k-1)!$ distinct cyclic orderings.

"$A_4$ has a subgroup of order 6 because it has 8 three-cycles." The eight $3$-cycles are exactly why it does not: any index-$2$ subgroup would contain all squares, hence all eight.

The exam's angle

1. Write the cycle decomposition first. Order, sign, class and centraliser all read off it.
2. Sign is $(-1)^{n-c}$. Count cycles including fixed points; it is faster and less error-prone than counting transpositions.
3. For conjugacy, compare cycle types — then check whether the question is in $A_n$.
4. For class sizes use $n!/\prod k^{a_k}a_k!$, and remember the centraliser order is $n!$ divided by the class size.
5. When counting classes, count partitions. No permutations need be written.

The night before

• Order $=$ lcm of cycle lengths. Sign $= (-1)^{n-c}$, $c$ counting fixed points.
• A $k$-cycle has sign $(-1)^{k-1}$: odd length is even, even length is odd.
• $\sigma(a_1\,a_2\cdots)\sigma^{-1} = (\sigma a_1\,\sigma a_2\cdots)$.
• Conjugate in $S_n$ $\iff$ same cycle type. Classes $\leftrightarrow$ partitions of $n$: $5$ for $S_4$, $7$ for $S_5$.
• Class size $= n!/\prod_k k^{a_k}a_k!$; centraliser order $= n!/(\text{class size})$.
• Classes split in $A_n$ when all cycle lengths are odd and distinct. $5$-cycles in $A_5$: $24$ becomes $12 + 12$.
• $A_5$ class equation $60 = 1+15+20+12+12$, and no sub-sum containing $1$ divides $60$ — hence simple.
• $S_n$ generated by transpositions, $A_n$ by $3$-cycles. $Z(S_n)$ trivial for $n\ge3$.