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…
Best Answer
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.