# Is this proof on the Sylow 5-subgroups of G 100% correct?

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 $$S_5.$$ • "... be expected in$S_4$"??? – user10354138 Nov 6 '18 at 21:30 • sorry that was a typo – excalibirr Nov 6 '18 at 21:31 •$S_5$nonabelian doesn't mean it cannot contain a proper normal subgroup. – user10354138 Nov 6 '18 at 21:33 • yes I had started to grow concerned over that bit being wrong after I'd posted it. is all the rest correct though ? – excalibirr Nov 6 '18 at 21:35 • 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. – user10354138 Nov 6 '18 at 21:38