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This is problem 26 from Grove's "Algebra."

Suppose $K$ is a conjugacy class in $S_n$ of cycle type $(k_1,...,k_n)$, and that $K \subseteq A_n$. If $\sigma \in K$ write $L$ for the conjugacy class of $\sigma$ in $A_n$.

If either $k_{2m} > 0$ or $k_{2m+1} > 1$ for some $m$ show that $L = K$.

I can show $L \subseteq K$ but not $K \subseteq L$. I don't know how to use the "$k_{2m} > 0$ or $k_{2m+1} > 1$" hypotheses. If $k_{2m} > 0$ for some $m$ then $\sigma \in A_n$ must have an even number of odd transpositions. Can I get a hint?

Thank you.

Edit: $k_m$ is the number of cycles of length m.

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Regarding the homework tag, it probably isn't necessary here. Mentioning that you're doing problems for practice is a fine idea, though. –  Dylan Moreland Jun 16 '12 at 17:06
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This recent question might be relevant, by the way. –  Dylan Moreland Jun 16 '12 at 17:08
    
If L⊆K and |L|=|K| then L=K, so you can use centralizers to calculate the size of L versus the size of K. Jim Belk's answer to math.stackexchange.com/questions/144686/… is similar to this. –  Jack Schmidt Jun 16 '12 at 17:10
    
math.stackexchange.com/questions/133599/… is a duplicate, but i like this version better and the old one has no answers. –  Jack Schmidt Jun 16 '12 at 17:13
    
@JackSchmidt It seems to be answered in the comments. –  Eugene Jun 16 '12 at 17:16

1 Answer 1

up vote 5 down vote accepted

Suppose you have $\sigma'\in K$; you want to show that $\sigma'\in L$.

Because $\sigma'\in K$ you have $\tau\sigma\tau^{-1}=\sigma'$ for some $\tau\in S_n$. Now if $\tau\in A_n$ then $\sigma'\in L$ immediately. The problem is if $\tau$ is odd and so not in $A_n$.

The idea is now that if we can find an odd $\rho$ such that $\rho\sigma\rho^{-1}=\sigma$, then we would have $\sigma'=\tau\rho\sigma\rho^{-1}\tau^{-1}$ with $\tau\rho\in A_n$, and we could conclude $\sigma'\in L$.

How do we find $\rho$? This is where the additional assumption on the cycle structure comes in. We know that either $\sigma$ has a cycle of even length, or $\sigma$ has two cycles of the same odd length. If the first is true, then (blah blah); otherwise the second is true and (blah blah). Can you take it from here?

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Ah, $k_m$ is the number of cycles of length $m$. That wasn't super clear to me. Great answer. –  Dylan Moreland Jun 16 '12 at 17:36
    
@DylanMoreland: I don't know if that is actually what it means, but it makes the claim true and is not too far-fetched an assumption ... –  Henning Makholm Jun 16 '12 at 17:38
    
@HenningMakholm: If $\sigma$ has cycle $\sigma_i$ of even length, let $\rho = \sigma_i$. Then $\rho$ is odd, and $\rho \sigma \rho^{-1} = \rho (\sigma_i...) \rho^{-1} = \rho (\sigma_i \rho^{-1}) ... = \sigma$ since $\sigma_i...$ are disjoint. –  Steven Li Jun 16 '12 at 21:21
    
@HenningMakholm: If $\sigma$ has cycles ${\sigma_i}_a$, ${\sigma_i}_b$ of same odd length $2m+1$, let $\rho = (a_1 b_1)(a_2 b_2)...(a_{2m+1} b_{2m+1})$ where ${\sigma_i}_a = (a_1...a_{2m+1})$ and ${\sigma_i}_b = (b_1...b_{2m+1})$. –  Steven Li Jun 16 '12 at 21:39
    
@StevenLi: Yes, exactly. –  Henning Makholm Jun 16 '12 at 21:58

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