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).