# Normal subgroup of group order 8

Prove that group of order 8 has hormal subgroup of order 4.

I know how to prove that if $G$ has a prime power order $p^n$ then it has a subgroup of order $p^m$ $\forall m \in \mathbb N: 0 \le m \le n$. And hence a group of order 8 has a subgroup of order 4, but how to prove that it has a normal subgroup of order 4?

Thanks for the help!

• Hint: what is the index of a subgroup of order $4$? – Ferra Dec 28 '14 at 17:31
• Hint: Consider the right and left cosets of a subgroup of order $4.$ – Geoff Robinson Dec 28 '14 at 17:31
• In fact all subgroups of index $p$ in a finite $p$-group are normal, although this is easier when $p = 2$. – KCd Dec 28 '14 at 17:36
• Index of a subgroup of order 4 equal 2, hence there are 2 left cosets $H$ and $aH$ and 2 right cosets $H$ and $Ha$. Hence $aH = Ha$ at it is normal subgroup. Is this true reasoning? – Stanislav Morozov Dec 28 '14 at 17:45
• @StanislavMorozov: Yes, that reasoning is correct, and you can conclde that $H$ is normal. – Geoff Robinson Dec 28 '14 at 17:48

You can prove a group $G$ of order $p^n$ has a normal subgroup of every order that divides $p^n$. To prove it use induction.

Since the center of a $p$-group is non trivial it has a subgroup of order $p$. call it $H$. Consider the projection $G\mapsto \frac{G}{H}$. Since $\frac{G}{H}$ has order $p^{n-1}$ use inductive hypothesis to find a normal subgroup of order $p^m$ where $m$ can go from $0$ to $1$. The preimage of that subgroup is normal and of order $p^{m+1}$ (use the firs iso theorem to prove this). Using this method you can find normal subgroups of all oders except $1$. But finding a normal subgroup of that order is not that hard.

This question can be solved many ways as in the comments, I prefered a simple one.

If $Z(G)=G$ we are done.

If $|Z(G)|=4$ we are done.

If $|Z(G)|=2$ then $G/Z(G)$ is abelian and has an normal subgroup of index $2$,So $G$ has.

And $Z(G)$ can not be trivial.