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Problem statement :

Let $G=S_5$ (it actually reads $|G|=S_5$ but this must be a misprint obviously). Using Sylow , how many 5-Sylow subgroups has G got? Is this consistent with the number of 5 cycles in G?

My proof:

we have $|G|=|S_5|=120=5\cdot 24$.

By Sylow's first theorem we now know that the order of the $5$-Sylow subgroups is 5.

By Sylow's 3rd theorem we know that the number of $5$-Sylow n_5=6.$

if $n_5=1$ then we also know that $|N_G(P)|=|G|$ but then this would imply that the normaliser contains every element in $G$ But of course $S_n$ is non-abelian for $n \geq 3$ so this can't be true.

We conclude that that $n_5=6$. So there are 6 $5$-Sylow -subgroups each with order $5$. These are distinct because the intersection is itself a subgroup and by Lagrange's theorem the intersection is either $1$ or $5$ and so must be $1$. so they intersect only at the identity. so any $5$-Sylow subgroup has order $5$ and contains the identity and $4$ $5$-cycles. Overall this means there are $(6\times 4=)\:24$ $5$-cycles, the same as would be expected in $S_5.$

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  • $\begingroup$ "... be expected in $S_4$"??? $\endgroup$ – user10354138 Nov 6 '18 at 21:30
  • $\begingroup$ sorry that was a typo $\endgroup$ – excalibirr Nov 6 '18 at 21:31
  • $\begingroup$ $S_5$ nonabelian doesn't mean it cannot contain a proper normal subgroup. $\endgroup$ – user10354138 Nov 6 '18 at 21:33
  • $\begingroup$ yes I had started to grow concerned over that bit being wrong after I'd posted it. is all the rest correct though ? $\endgroup$ – excalibirr Nov 6 '18 at 21:35
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    $\begingroup$ The rest is OK (athough I wouldn't call calculating the order of a Sylow $p$ the content of the first Sylow theorem). To conclude $n_5=6$, look at the subgroup $A_5$ which is simple. $\endgroup$ – user10354138 Nov 6 '18 at 21:38

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