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I'm attempting this homework problem, and I'm not sure where to start. Here is the problem and how what I've got so far.

Let $p$ be a prime number. What is the least positive integer $n$ such that $S_n$ (The symmetric group on $n$ objects) has a subgroup of order $p^2$.

I think whatever approach is taken, it will probably use Lagrange's theorem. $S_n$ is a finite group with order $n!$, so if it has a subgroup of order $p^2$ then $p^2 | n!$.

But the converse of Langrange's theorem is not true generally. We can't say that if $p^2|n!$ then $S_n$ has a subgroup of order $p^2$. Maybe if I had some insight into WHEN the converse of Lagrange's theorem is true I would know better what to do next.

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  • $\begingroup$ If $p^k$ divides $n$ for $p$ prime and $k$ a non-negative integer, then every group of order $n$ has at least one subgroup of order $p^k$. CLT holds for prime powers. $\endgroup$
    – Jack Schmidt
    Commented May 2, 2013 at 5:07

2 Answers 2

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If $p^2$ divides $n!$ then $n\geq 2p$, and then you can find two disjoint $p$ cycles in $S_n$: they generate a group of order $p^2$.

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  • $\begingroup$ That simple...+1. $\endgroup$
    – Julien
    Commented May 2, 2013 at 5:06
  • $\begingroup$ So if $n = 2p$ we can find two disjoint subgroups of order $p$? I can see how we find just one, since if $n = 2p$ then $p | n | n!$. $\endgroup$
    – objectivesea
    Commented May 2, 2013 at 6:04
  • $\begingroup$ Let me be more explicit: the cycles $(1,2,\dots,p)$ and $(p+1,p+1,\dots,2p)$ are in $S_n$ if $n\geq2p$ (and this last condition is equivalnt to $p^2\mid n!$) and they generate a group of order $p^2$. I don't know if that answers your comment, for I don't understand it, really :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 2, 2013 at 6:11
  • $\begingroup$ @MarianoSuárez-Álvarez This can be generalized to $p^k$ for any $k \geq 1$, right? (if $p$ is still prime of course) $\endgroup$
    – Evariste
    Commented Sep 28, 2018 at 21:31
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If $n=2p$, then $p^2|n!$ but $p^3\nmid n!$. Now use Sylow's First Theorem. As @Mariano points out, if $n<2p$, then $p^2\nmid n!$ so no such subgroup exists.

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