# Splitting conjugacy classes in alternating group

Re­call that in the sym­met­ric group $$S_n$$, the no­tion of “con­ju­gacy class” co­in­cides with that of “has the same cy­cle type” (proof). It turns out to be the case that when we de­scend to the al­ter­nat­ing group $$A_n$$, the con­ju­gacy classes nearly all re­main the same; the ex­cep­tion is those classes in $$S_n$$ which have cy­cle type all odd and all dis­tinct. Those classes split into ex­actly two classes in $$A_n$$.

# Proof

## Con­ju­gate im­plies same cy­cle type

This di­rec­tion of the proof is iden­ti­cal to the same di­rec­tion in the proof of the cor­re­spond­ing re­sult on sym­met­ric groups.

## Con­verse: split­ting condition

Re­call be­fore we start that an even-length cy­cle can only be writ­ten as the product of an odd num­ber of trans­po­si­tions, and vice versa.

The ques­tion is: is ev­ery $$\tau \in A_n$$ with the same cy­cle type as $$\sigma \in A_n$$ con­ju­gate (in $$A_n)$$ to $$\sigma$$? If so, then the con­ju­gacy class in $$A_n$$ of $$\sigma$$ is just the same as that of $$\sigma$$ in $$S_n$$; if not, then the con­ju­gacy class in $$S_n$$ must break into two pieces in $$A_n$$, namely $$\{ \rho \sigma \rho^{-1} : \rho \ \text{even} \}$$ and $$\{ \rho \sigma \rho^{-1} : \rho \ \text{odd} \}$$. (Cer­tainly these are con­ju­gacy classes: they only con­tain even per­mu­ta­tions so they are sub­sets of $$A_n$$, while ev­ery­thing in ei­ther class is con­ju­gate in $$A_n$$ be­cause the defi­ni­tion only de­pends on the par­ity of $$\rho$$.)

Let $$\sigma = c_1 \dots c_k$$, and $$\tau = c_1' \dots c_k'$$ of the same cy­cle type: $$c_i = (a_{i1} \dots a_{i r_i})$$, $$c_i' = (b_{i1} \dots b_{i r_i})$$.

Define $$\rho$$ to be the per­mu­ta­tion which takes $$a_{ij}$$ to $$b_{ij}$$, but oth­er­wise does not move any other let­ter. Then $$\rho \sigma \rho^{-1} = \tau$$, so if $$\rho$$ lies in $$A_n$$ then we are done: $$\sigma$$ and $$\tau$$ are con­ju­gate in $$A_n$$.

That is, we may as­sume with­out loss of gen­er­al­ity that $$\rho$$ is odd.

### If any of the cy­cles is even in length

Sup­pose with­out loss of gen­er­al­ity that $$r_1$$, the length of the first cy­cle, is even. Then we can rewrite $$c_1' = (b_{11} \dots b_{1r_1})$$ as $$(b_{12} b_{13} \dots b_{1 r_1} b_{11})$$, which is the same cy­cle ex­pressed slightly differ­ently.

Now $$c_1'$$ is even in length, so it is an odd per­mu­ta­tion (be­ing a product of an odd num­ber of trans­po­si­tions, $$(b_{1 r_1} b_{11}) (b_{1 (r_1-1)} b_{11}) \dots (b_{13} b_{11})(b_{12} b_{11})$$). Hence $$\rho c_1'$$ is an even per­mu­ta­tion.

But con­ju­gat­ing $$\tau$$ by $$\rho c_1'$$ yields $$\sigma$$:

$$\sigma = \rho \tau \rho^{-1} = \rho c_1' (c_1'^{-1} \tau c_1') c_1'^{-1} \rho^{-1}$$
which is the re­sult of con­ju­gat­ing $$c_1'^{-1} \tau c_1' = \tau$$ by $$\rho c_1'$$.

(It is the case that $$c_1'^{-1} \tau c_1' = \tau$$, be­cause $$c_1'$$ com­mutes with all of $$\tau$$ ex­cept with the first cy­cle $$c_1'$$, so the ex­pres­sion is $$c_1'^{-1} c_1' c_1' c_2' \dots c_k'$$, where we have pul­led the fi­nal $$c_1'$$ through to the be­gin­ning of the ex­pres­sion $$\tau = c_1' c_2' \dots c_k'$$.)

Sup­pose $$\sigma = (12)(3456), \tau = (23)(1467)$$.

We have that $$\tau$$ is taken to $$\sigma$$ by con­ju­gat­ing with $$\rho = (67)(56)(31)(23)(12)$$, which is an odd per­mu­ta­tion so isn’t in $$A_n$$. But we can rewrite $$\tau = (32)(1467)$$, and the new per­mu­ta­tion we’ll con­ju­gate with is $$\rho c_1' = (67)(56)(31)(23)(12)(32)$$, where we have ap­pended $$c_1' = (23) = (32)$$. It is the case that $$\rho$$ is an even per­mu­ta­tion and hence is in $$A_n$$, be­cause it is the re­sult of mul­ti­ply­ing the odd per­mu­ta­tion $$\rho$$ by the odd per­mu­ta­tion $$c_1'$$.

Now the con­ju­ga­tion is the com­po­si­tion of two con­ju­ga­tions: first by $$(32)$$, to yield $$(32)\tau(32)^{-1} = (23)(1467)$$ (which is $$\tau$$ still!), and then by $$\rho$$. But we con­structed $$\rho$$ so as to take $$\tau$$ to $$\sigma$$ on con­ju­ga­tion, so this works just as we needed. <div><div>

## If all the cy­cles are of odd length, but some length is repeated

Without loss of gen­er­al­ity, sup­pose $$r_1 = r_2$$ (and la­bel them both $$r$$), so the first two cy­cles are of the same length: say $$\sigma$$‘s ver­sion is $$(a_1 a_2 \dots a_r)(c_1 c_2 \dots c_r)$$, and $$\tau$$’s ver­sion is $$(b_1 b_2 \dots b_r)(d_1 d_2 \dots d_r)$$.

Then define $$\rho' = \rho (b_1 d_1)(b_2 d_2) \dots (b_r d_r)$$. Since $$r$$ is odd and $$\rho$$ is an odd per­mu­ta­tion, $$\rho'$$ is an even per­mu­ta­tion.

Now con­ju­gat­ing by $$\rho'$$ is the same as first con­ju­gat­ing by $$(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$$ and then by $$\rho$$.

But con­ju­gat­ing by $$(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$$ takes $$\tau$$ to $$\tau$$, be­cause it has the effect of re­plac­ing all the $$b_i$$ by $$d_i$$ and all the $$d_i$$ by $$b_i$$, so it sim­ply swaps the two cy­cles round.

Hence the con­ju­ga­tion of $$\tau$$ by $$\rho'$$ yields $$\sigma$$, and $$\rho'$$ is in $$A_n$$.

Sup­pose $$\sigma = (123)(456), \tau = (154)(237)$$.

Then con­ju­ga­tion of $$\tau$$ by $$\rho = (67)(35)(42)(34)(25)$$, an odd per­mu­ta­tion, yields $$\sigma$$.

Now, if we first perform the con­ju­ga­tion $$(12)(53)(47)$$, we take $$\tau$$ to it­self, and then perform­ing $$\rho$$ yields $$\sigma$$. The com­bi­na­tion of $$\rho$$ and $$(12)(53)(47)$$ is an even per­mu­ta­tion, so it does line in $$A_n$$. <div><div>

## If all the cy­cles are of odd length, and they are all distinct

In this case, we are re­quired to check that the con­ju­gacy class does split. Re­mem­ber, we started out by sup­pos­ing $$\sigma$$ and $$\tau$$ have the same cy­cle type, and they are con­ju­gate in $$S_n$$ by an odd per­mu­ta­tion (so they are not ob­vi­ously con­ju­gate in $$A_n$$); we need to show that in­deed they are not con­ju­gate in $$A_n$$.

In­deed, the only ways to rewrite $$\tau$$ into $$\sigma$$ (that is, by con­ju­ga­tion) in­volve tak­ing each in­di­vi­d­ual cy­cle and con­ju­gat­ing it into the cor­re­spond­ing cy­cle in $$\sigma$$. There is no choice about which $$\tau$$-cy­cle we take to which $$\sigma$$-cy­cle, be­cause all the cy­cle lengths are dis­tinct. But the per­mu­ta­tion $$\rho$$ which takes the $$\tau$$-cy­cles to the $$\sigma$$-cy­cles is odd, so is not in $$A_n$$.

More­over, since each cy­cle is odd, we can’t get past the prob­lem by just cy­cling round the cy­cle (for in­stance, by tak­ing the cy­cle $$(123)$$ to the cy­cle $$(231)$$), be­cause that in­volves con­ju­gat­ing by the cy­cle it­self: an even per­mu­ta­tion, since the cy­cle length is odd. Com­pos­ing with the even per­mu­ta­tion can’t take $$\rho$$ from be­ing odd to be­ing even.

There­fore $$\tau$$ and $$\sigma$$ gen­uinely are not con­ju­gate in $$A_n$$, so the con­ju­gacy class splits.

Sup­pose $$\sigma = (12345)(678), \tau = (12345)(687)$$ in $$A_8$$.

Then con­ju­ga­tion of $$\tau$$ by $$\rho = (87)$$, an odd per­mu­ta­tion, yields $$\sigma$$. Can we do this with an even per­mu­ta­tion in­stead?

Con­ju­gat­ing $$\tau$$ by any­thing at all must keep the cy­cle type the same, so the thing we con­ju­gate by must take $$(12345)$$ to $$(12345)$$ and $$(687)$$ to $$(678)$$.

The only ways of $$(12345)$$ to $$(12345)$$ are by con­ju­gat­ing by some power of $$(12345)$$ it­self; that is even. The only ways of tak­ing $$(687)$$ to $$(678)$$ are by con­ju­gat­ing by $$(87)$$, or by $$(87)$$ and then some power of $$(678)$$; all of these are odd.

There­fore the only pos­si­ble ways of tak­ing $$\tau$$ to $$\sigma$$ in­volve con­ju­gat­ing by an odd per­mu­ta­tion $$(678)^m(87)$$, pos­si­bly alongside some pow­ers of an even per­mu­ta­tion $$(12345)$$; there­fore to get from $$\tau$$ to $$\sigma$$ re­quires an odd per­mu­ta­tion, so they are in fact not con­ju­gate in $$A_8$$. <div><div>

# Example

In $$A_7$$, the cy­cle types are $$(7)$$, $$(5, 1, 1)$$, $$(4,2,1)$$, $$(3,2,2)$$, $$(3,3,1)$$, $$(3,1,1,1,1)$$, $$(1,1,1,1,1,1,1)$$, and $$(2,2,1,1,1)$$. The only class which splits is the $$7$$-cy­cles, of cy­cle type $$(7)$$; it splits into a pair of half-sized classes with rep­re­sen­ta­tives $$(1234567)$$ and $$(12)(1234567)(12)^{-1} = (2134567)$$.

Parents:

• Alternating group

The al­ter­nat­ing group is the only nor­mal sub­group of the sym­met­ric group (on five or more gen­er­a­tors).