[Math] Sylow’s Theorem Application. Prove $G$ is Abelian.

Question

Assume that $|G|=5^27^2$. Determine the possibilities for $n_5, n_7$ and determine what can be concluded in each case about the $5$-Sylow subgroups and the $7$-Sylow subgroups and prove that $G$ is Abelian.

Proof: Assume that |G|=$5^27^2$. By Sylow's Theorem,

Then $n_5 \equiv 1\pmod 5$ and $n_5 \mid 49$. Thus $n_5 = 1$.

And $n_7 \equiv 1\pmod 7$ and $n_7 \mid 25$. Thus $n_7 = 1$.

Hence $P_5 \lhd G$ and $P_7 \lhd G$.

I am stuck on how to prove $G$ is abelian.

Answer 1

Do you know that if $p$ is a prime number, then every group of order $p^2$ is abelian? Use that and finish the proof with this one weird tip:

• If $M \trianglelefteq G$ and $N \trianglelefteq G$ and $M \cap N = 1$, then $mn = nm$ for all $m \in M$ and $n \in N$.

(you should prove this if you haven't already)