# 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.

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Hint: If all elements have order $2$ the group is abelian. If there is an element of order $4$, it generates a subgroup of index $2$, which is then normal. – Tobias Kildetoft Apr 30 '13 at 15:02
Here is a silly answer that is wrong: The number of Sylow 2-subgroups is a divisor of 8 that is congruent to 1 mod 2, so the number of Sylow 2-subgroups is exactly 1. Since there is only one Sylow 2-subgroup, it must be normal. $\square$ – Jack Schmidt Apr 30 '13 at 15:06
@JackSchmidt Very nice. niladri: Note that the method I used here was intentionally very elementary. In general, one can show that no group of order $p^n$ for a prime $p$ is simple (they are in fact nilpotent), but this requires a bit more argument. – Tobias Kildetoft Apr 30 '13 at 15:10
You can also use the classification of simple groups and check that no group in that list has order $8$.... Just kidding, sorry couldn't resist.... – N. S. Apr 30 '13 at 15:20
I should perhaps add that what I wrote about groups of order $p^n$ is not completely correct, as any group of prime order is of course simple by Langrange's theorem. – Tobias Kildetoft Apr 30 '13 at 15:22

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. $\square$
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.