# Subgroups of a finite, normal, cyclic group.

Let $G$ be a group and $H$ be a normal subgroup of $G$. Assume that $H$ is finite and cyclic. Show that every subgroup of $H$. is also normal in $G$.$$I am trying to prove this problem, but absolutely cannot make any progress. I know that H is generated by an element a\in G with the order of H being the order of a. I also know that any subgroup K of H will be cyclic as well. I cannot seem to make any further progress from this.. any help would be great. Thank you. ## 3 Answers Let a be a generator of H. Then any subgroup K of H is generated by a^k for some k. For any g \in G, we have ga^kg^{-1} = (gag^{-1})^k  and gag^{-1} \in H since H is normal. Thus gag^{-1} = a^l for some l. Thus$$ga^kg^{-1} = (gag^{-1})^k = (a^l)^k = (a^k)^l \in K since $K$ is generated by $a^k$. It follows that $K$ is normal.

• Why is $ga^kg^-1 = (gag^-1)^k)$? – Sank Nov 3 '16 at 20:37
• $ga^2g^{-1}=gag^{-1}gag^{-1} = (gag^{-1})^2$. Use induction. – user348749 Nov 3 '16 at 23:54

There is a generalization here. A subgroup $K$ of a group $H$ is called characteristic, if for any $\alpha \in Aut(H)$, $\alpha[K]=K$, and one writes $K$ char $H$. Note that $K$ is necessarily normal, since the inner automorphisms of $H$ fix $K$. Examples of characteristic subgroups are the commutator subgroup $H'$ and the center $Z(H)$.
Now, one can show in general that $K$ char $H \unlhd G$, implies $K \unlhd G$. So the thing you have to show is that all subgroups of a cyclic group are characteristic. And dear user282639, I leave that to you.

As H is cyclic implies H is abelian and that implies every subgroup of H is normal in H but as H is normal in G which clearly implies every subgroup of H is normal in G.