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Let $n_{k}$ be the number of elements of order $k$ in the group $A_{n}$ (Alternating group of degree $n$). Also let $p$ be the greatest prime divisor in $\pi (A_{n})$. We know that $n_{k}(A_{n})=\sum |cl_{A_{n}}(x_{i})|$ such that $|x_{i}|=k$. My question is: Is there any $k\neq p$ such that $p\mid (1+n_{k})$?

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What is $\pi(A_n)$? – Chris Eagle Nov 5 '12 at 17:10
For an integer $n$, $\pi(n)$ is the set of all prime divisors of $n$. $\pi(G)=\pi(|G|)$ is the set of primes $p$ such that $G$ contains an element of order $p$. – user2132 Nov 5 '12 at 17:14
But $A_n$ isn't an integer? – joriki Nov 5 '12 at 17:55
Also, in case you mean $|A_n|$, you'll need $n\ge3$ for $|A_n|$ to have any prime divisors. – joriki Nov 5 '12 at 18:03
up vote 2 down vote accepted

As discussed in the comments, I'll assume that $p$ is the greatest prime divisor of $|A_n|$.

Yes, there is such a $k\ne p$. To find one without help from our electronic friends, we need to understand why there are no examples for low $n$. Consider the orbits of $A_n$ under conjugation by some $p$-cycle $\sigma$. Their size divides $p$ and is therefore either $1$ or $p$. For a permutation $\pi$ to be self-conjugate under $\sigma$, $\sigma$ must permute cycles of equal length among themselves. It can only do so if $\pi$ is a power of $\sigma$ (including the identity) on the range of $\sigma$. As long as $n\lt p+3$, this is only the case if $\pi$ is a power of $\sigma$ (including the identity). Thus all orbits except for the two containing the powers of $\sigma$ have size $p$, and these two have size $1$ for the identity and $p-1$ for the remaining powers of $\sigma$, leading to $p\mid(1+n_p)$.

However, once $n\ge p+3$, which first happens for $n=10$, this is no longer the case, and we can try to construct a counterexample. To do so systematically, take $\pi$ a power of $\sigma$ on the range of $\sigma$, and add another odd cycle of length $l$. There are $(l-1)!$ different possibilities for that cycle, and $p-1$ non-trivial powers of $\sigma$, so we want $(l-1)!\equiv1\bmod p$. By a happy coincidence, for $l=5$ we have $(l-1)!=24\equiv1\bmod23$, and $23$ happens to be the first prime followed by a prime gap of $6$, so for $n=23+5=28$ we still have $p=23$. The only elements of $A_{28}$ of order $k=23\cdot5$ are those consisting of a $23$-cycle and a $5$-cycle, and they number $\binom{28}522!4!\equiv1\cdot(-1)\cdot1\equiv-1\bmod23$.

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Thank you for your nice answer. – user2132 Nov 6 '12 at 17:46
@user2132: You're welcome! – joriki Nov 6 '12 at 21:09

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