Show that a group of order $180$ is not simple.

What i deduced is that $n_5=1,6$ or $36$. We are done if $n_5=1$.

If $n_5=36$ we $N_G(P)=P$ for any Sylow $5$-subgroup P as $|N_G(P)|=\frac{180}{36}=5$ and $P$ is abelian cyclic so by Burnside p-complement theorem there exist a normal subgroup of order $36$ which is complement to $P$ , fine, and we are done, not simple.

For $n_5=6$, $|N_G(P)|=\frac{180}{6}=30$. From here I can see that by $N/C$- theorem $|N_G(P)/C_G(P)|\le 2$. In this case I am stuck, Any help please.

Thanks in advance!

Solution 1:

Assume $G$ is simple and that $n_5=6$, so index$[G:N_G(P)]=6$, $P \in Syl_5(G)$. But then $G$ embeds homomorphically into $A_6$ (note that core$_G(N_G(P))=1$, since $G$ is simple!). But index$[A_6:G]=360/180=2$, implying $G$ is normal in $A_6$, which contradicts the simplicity of $A_6$.