# Prove a group that has a normal subgroup isomorphic to $D_8$ has a non-trivial center

Let $G$ be a group which has a normal subgroup isomorphic to $D_8$. Prove that $G$ has a non trivial center.

So, given $g\in G$, $h\in D_8$ $ghg^{-1}\in D_8$. So I tried to prove that there is an element (not equal the identity) $h\in D_8$ such that $ghg^{-1}=h$, but no success so far. I used trial and error method. But did not try all because there are lot. My question is, is it possible to find such element? Or else what other method works here?

• The first step is to realize that $D_8$ has non-trivial center. Prove this first. (Although this does not imply yet the claim.) – Martin Brandenburg Jan 3 '16 at 18:13

## 2 Answers

Hint: if $N$ is a normal subgroup of $G$, and $|N|=2$, then $N \subseteq Z(G)$.

• Great answer! thanks. – Extremal Jan 3 '16 at 20:00
• You are welcome! – Nicky Hekster Jan 3 '16 at 20:50

Note that the centre of a group $D_8$ is a characteristic subgroup of $D_8$. Hence any automorphism of $D_8$ (e.g., those that are inner automorphisms of $G$) leave it invariant. Hence you should determine the centre of $D_8$ itself - and then verify that it has only few automorphisms.