# Conjugacy class splitting in double cover of alternating, symmetric group

Let $n \ge 4$ be a natural number.

Consider the quotient map from the double cover $2 \cdot A_n$ of the alternating group $A_n$ to $A_n$. For any conjugacy class in $A_n$ of size $r$, the inverse image in $2 \cdot A_n$ has size $2r$. This inverse image is either a single conjugacy class or "splits" as a union of two conjugacy classes each of size $r$. Which case occcurs depends on the original conjugacy class we chose.

I want a combinatorial criterion (see Note 2) on the cycle type of the conjugacy class in $A_n$ that helps determine which of these cases occur (actually, there may be one or two conjugacy classes in $A_n$ of a given cycle type, but even if there are two, they will behave the same way).

I'm also interested in the corresponding question for the double cover of $S_n$. Note that the double cover is not unique (there are two possibilities for each $S_n$). However, the behavior for fibers over conjugacy classes that I'm interested in should not depend on the choice of double cover (if my understanding is correct -- I haven't worked out a formal proof).

Note 1: I'm including the case of $A_n$ for $n = 4$ just so people know what I am talking about:

$A_4$ has conjugacy classes of sizes 1,3,4,4 (the two conjugacy classes of size 4 fuse in $S_4$ and correspond to cycle type 3 + 1, whereas the conjugacy class of size 3 corresponds to cycle type 2 + 2, with representative permutation $(1,2)(3,4)$).

Its double cover is isomorphic to $SL(2,3)$ and the quotient map can be viewed as a quotient map $SL(2,3) \to PSL(2,3)$. The fibers over the conjugacy classes of sizes 1,4,4, each split in two conjugacy classes. The fiber over the conjugacy class of size 3 remains a single conjugacy class of size 6, i.e., the fiber over this conjugacy class does not split.

Note 2: By "combinatorial criterion" I mean a criterion like the criterion we have for a conjugacy class of even permutations in $S_n$ to split in $A_n$: it splits if there is any cycle of even length or two cycles of equal odd length.

So, I want a criterion that looks at the cycle type as an unordered integer partition, then uses the combinatorial and number-theoretic structure of that partition to determine splitting in the double cover.

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I believe the answer is that the class of an element of $g \in A_n$ becomes a single class when lifted to the 2-fold central covering group of $A_n$ if and only if $g$ has even order and $g$ has (at least) two cycles of the same length (including cycles of length 1).

The class of an element of $g \in S_n$ lifts to a single class in either of the two covering groups of $S_n$ if and only if either the condition for $g \in A_n$ holds, or $g$ has an even number of cycles of even length.

Let $\hat{S}$ and $\hat{A}$ be covering groups of $S_n$ and $A_n$. Let $z$ be the central element of order 2 in $\hat{S}$. To avoid too many hats, I will denote the inverse image in $\hat{S}$ of $g \in S_n$ also by $g$.

I believe that to study these covering groups, all you really need to know is that if $g$ and $h$ are disjoint transpositions in $S_n$, such as $(1,2)$ and $(3,4)$, then $[g,h] = z$ in $\hat{S}$. Also, for two commuting pairs of transpositions in $A_4$, like $g=(1,2)(3,4)$ and $h = (1,3)(2,4)$, we have $[g,h]=z$ in $\hat{A}$.

Clearly, if $g$ has odd order $k$ in $A_n$ or $S_n$, then its class lifts to two classes in $\hat{A}$ or $\hat{S}$, one of order $k$ and one of order $2k$, so we only need consider elements $g$ of even order.

If $g$ has two cycles of equal even length, such as $(1,2,3,4)(5,6,7,8)$ (+ other cycles) then we can take $h = (1,5)(2,6)(3,7)(4,8)$ and get $[g,h] = z$ in $\hat{A}$, so the class of $g$ lifts to a single class in $\hat{A}$. I had some difficulty proving that, but if it was false, then you could show that the centralizer of $h$ in $\hat{A}$ mapped on to the full centralizer of $h$ in $A_n$, which is not true, because the image of the centralizer does not contain $(1,2)(5,6)$.

On the other hand, if $g$ has even order and has two cycles of the same odd length, such as $g = (1,2)(3,4,5,6)(7,8,9)(10,11,12)$ with $n=12$, then we can take $h = (1,2)(7,10)(8,11)(9,12)$ and get $[g,h] = z$ in $\hat{A}$.

But if all cycles of $g$ have different lengths, then elements $h$ of the centralizer of $g$ in $A_n$ are made up of powers of those cycles, and since $h$ must have an even number of cycles of even length, we always get $[g,h]=1$ in $\hat{A}$.

However, in $S_n$, if $g$ has an even number of cycles of even length, like $g=(1,2)(3,4,5,6)$ with $n=6$, then we can take $h=(1,2)$ and get $[g,h]=z$ in $\hat{S}$.

Let me know if I have got something wrong, or if you would like more detail!.

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Thanks a lot! This seems correct and is exactly what I was looking for! However, I haven't gone through all your details carefully. I will look at it more carefully later and let you know if I have further questions. – Vipul Naik May 28 '12 at 3:37