Prove that a group of order $60$ with more than one Sylow-$5$ subgroup is simple.

I have shown that there are $6$ Sylow-$5$ subgroups, and I may use the fact that $A_5$ is simple, so I just need to show that the only possible group this can be is $A_5$. I am not sure what to do next though...

  • $\begingroup$ There may be different ways of doing this, but personally I would ignore the idea of using the simplicity of $A_5$ and prove simplicity directly, by considering orders of possible normal subgroups, and showing that whenever there is a proper normal subgroup there must be a single Sylow $5$-subgroup. $\endgroup$ – Derek Holt Feb 8 '18 at 8:51

We first prove that a group of order $30$ must have a normal Sylow $5$-subgroup. The other possibility is that it has $6$, but then there would only be $6$ elements not having order $5$, so it would have a normal Sylow $3$-subgroup $N$, then $G/N$ would have a normal Sylow $5$-subgroup and hence so would $G$, contradiction.

Now suppose that $|G| = 60$, and let $1 < N \lhd G$.

Case 1. If $5$ divides $|N|$, then by Sylow's theorem and the result above $N$ has a normal Sylow $5$-subgroup, which must also be normal in $G$.

Case 2. Otherwise, by Sylow or by the result above $G/N$ must have a normal Sylow $5$-subgroup $P/N$, and so Case 1 applies to $P \lhd G$ except when $|N|=12$ and $P=G$.

Case 3. If $|N|=12$, then by Sylow and counting, $N$ must have either a normal Sylow $2$-subgroup of a normal Sylow $3$-subgroup, which is then normal in $G$, and we back in Case 2.

So if $G$ has $6$ Sylow $5$-subgroups then it must be simple.

  • $\begingroup$ Why does G/N having a normal sylow-5 subgroup imply that G has a normal Sylow-5 subgroup? $\endgroup$ – user85798 Feb 9 '18 at 1:03
  • $\begingroup$ @bwv869 Last sentence first paragraph: I'm guessing he is using this fact: if $K$ is characteristic in $H$ and $H$ is normal in $G$, then $K$ is normal in $G$. Here, $P/N$ is normal in $G/N$ with $|P/N|=5$. So $P$ is normal in $G$ with $|P|=15$. Hence $P$ has a normal Sylow $5$-subgroup $K$, with $|K|=5$. Then $K$ is characteristic in $P$. So $K$ is normal in $G$. Hence $G$ has a normal Sylow $5$-subgroup. $\endgroup$ – Delong Feb 9 '18 at 4:14
  • $\begingroup$ @Delong yes that's right. $\endgroup$ – Derek Holt Feb 9 '18 at 8:18

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