Groups of order 8 are not simple Show that any group of order 8 is not a simple group. 
I know that $\mathbb{Z}_8$, $\mathbb{Z}_2\times \mathbb{Z}_4$, $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$, $Q_8$, $D_4$ are not simple. But I am unable to prove it generally. 
Please help.
 A: Here is another silly proof, but this one is probably right.
By Cauchy's theorem (or Lagrange) a group $G$ of order 8 contains a subgroup $H$ of order 2. Consider the homomorphism from $G$ to $\operatorname{Sym}(4)$ given by number the 4 cosets of $H$ in $G$, and letting $G$ act as multiplication on the cosets. The image is transitive (moves all the points around) so the kernel must be smaller than $G$. If $G$ is simple, then the kernel has to be the identity. The first isomorphism theorem shows that $G$ is isomorphic to a subgroup $\operatorname{Sym}(4)$, but every subgroup of order 8 in $\operatorname{Sym}(4)$ is a Sylow 2-subgroup, and so dihedral of order 8. Since dihedral groups of order 8 are already known to be non-simple, we are done. $\square$
Usually instead of the cosets of a subgroup of order 2, one uses a bigger subgroup, but 8 is so small, this just works anyways.
A: This group is a $p$-group with $p=2$, so the center of the group, $Z(G)$, is nontrivial, and $Z(G)$ is normal in $G$ (true for any group). Moreover, $Z(G)=G$ $\iff$ $G$ is abelian, in which case any subgroup of order $2$, for this specific problem, is normal. If $Z(G)\neq G$, then $Z(G)$ is a nontrivial proper normal subgroup. 
