Recall that conjugacy classes in the symmetric group are determined solely by cycle type. In the alternating group , however, this is not always true. A single conjugacy class in that is contained in may split into two distinct classes when considered as a subset of . For example, in , and are conjugate, since
but these two are not conjugate in (note that ).
Note in particular that the fact that conjugacy in is determined by cycle type means that if then all of its conjugates in also lie in .
The following theorem fully characterizes the behavior of conjugacy classes in :
A conjugacy class in splits into two distinct conjugacy classes under the action of if and only if its cycle type consists of distinct odd integers. Otherwise, it remains a single conjugacy class in .
Thus, for example, in , the elements of the conjugacy class of are all conjugate in , while the elements of the conjugacy class of split into two distinct conjugacy classes in since there are two cycles of length . Similarly, any conjugacy class containing an even-length cycle, such as , splits in .
We will prove the above theorem by proving the following statements:
If , then the elements of the conjugacy class of in (which is just all elements of the same cycle type as ) are conjugate in if and only if commutes with some odd permutation.
does not commute with an odd permutation if and only if the cycle type of consists of distinct odd integers.
Throughout, we will denote by the conjugacy class of under the action of .
To prove the first statement, note that conjugacy is a transitive action. By the theorem that orbits of a normal subgroup are equal in size when the full group acts transitively, we see that if , then splits into classes under the action of (recall that , the centralizer of , is simply the stabilizer of under the conjugation action of on itself). But since is either or , we see that the conjugacy class of either remains a single class in or splits into two classes.
Note also that the elements of are all conjugate in if and only if , which happens if and only if , which in turn is the case if and only if some odd permutation is in the centralizer of , which means precisely that commutes with some odd permutation. This proves the second statement.
To prove the third statement, suppose first that does not commute with an odd permutation. Clearly commutes with any cycle in its own cycle decomposition, so if contains a cycle of even length, that is an odd permutation with which commutes. So must consist solely of [disjoint] cycles of odd length. If two of these cycles have the same length, say and , then
To prove the converse, we show that if the cycles in the cycle decomposition all have distinct lengths, then commutes precisely with the group generated by its cycles. It follows then that if all the distinct lengths are odd, then commutes only with these permutations, which are all even. Choose with distinct cycle lengths in its cycle decomposition, and suppose that commutes with some element . Conjugation preserves cycle length, so since commutes with and has all its cycles of distinct lengths, each cycle in must commute with each cycle in individually.
Now, choose a nontrivial cycle of , and choose such that moves (we can do this, since can have at most one cycle of length and the cycle length of is greater than ). Let be the cycle of containing . Then commutes with since commutes with , so is in the centralizer of , and it is not disjoint from . But the centralizer of a -cycle consists of products of powers of and cycles disjoint from . Thus is a power of . So each cycle in is a power of a cycle in , and we are done.
|Date of creation||2013-03-22 17:18:04|
|Last modified on||2013-03-22 17:18:04|
|Last modified by||rm50 (10146)|