6
$\begingroup$

I'd appreciate comments on the validity of these attempted proofs. Thanks.

Let $a$ be an $n$-cycle in $S_n$.

a) Show that the centralizer of $a$ in $S_n$ is $\langle a \rangle$.

b) Assume that $n$ is odd. Show that $A_n$ contains an $n$-cycle $b$ that is not conjugate to $a$ in $A_n$.

a) Since the number of $k$-cycles in $S_n$ is given by $\frac{n!}{(n-k)!k}$, the number of $n$-cycles in $S_n$ is $(n-1)!$. Since two elements of $S_n$ are conjugate iff they have the same cycle type, $|\mathrm{cl}(a)|=(n-1)!$ (where $\mathrm{cl}(a)$ is the conjugacy class of $a$). But $|\mathrm{cl}(a)|=|S_n:C_{S_n}(a)|$, or $|C_{S_n}(a)|=\frac{|S_n|}{|\mathrm{cl}(a)|}=\frac{n!}{(n-1)!}=n$. Since $\langle a \rangle \le C_{S_n}(a)$, it follows that $\langle a \rangle = C_{S_n}(a)$.

b) since $n$ is odd, all $n$-cycles are contained in $A_n$, so $A_n$ contains $(n-1)!\,\,$ $n$-cycles. Since, for $a \in S_n$, $C_{A_n}(a)=C_{S_n}(a) \cap A_n = \langle a \rangle \cap A_n$, we have $|\mathrm{cl}_{A_n}(a)|=|A_n:C_{A_n}(a)|=\frac{\frac{1}{2}n!}{n}=\frac{n!}{2n}=\frac{1}{2}(n-1)!$, and so $a$ cannot be conjugate to all $(n-1)!$ $n$-cycles in $A_n$ and there exists such an element $b$.

$\endgroup$
  • $\begingroup$ Yes, those are both correct. And clearly written, too. $\endgroup$ – Christopher Feb 5 '14 at 17:22
  • $\begingroup$ Great, thanks.$\,$ $\endgroup$ – Alex Petzke Feb 6 '14 at 1:46
  • $\begingroup$ Perhaps you made a little mistake when you wrote: $C_{A_n}(a) =C_{A_n} \cap A_n$. Did you mean $C_{S_n}(a) =C_{A_n} \cap A_n$? $\endgroup$ – Jonas Gomes Mar 28 '14 at 14:29
  • $\begingroup$ Yes, this should really be a comment. But I did make a correction, though it is not what you suggested. It should be $C_{A_n}(a)=C_{S_n}(a) \cap A_n$, not $C_{A_n}(a)=C_{A_n}(a) \cap A_n$. $\endgroup$ – Alex Petzke Mar 28 '14 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.