# If $N$ is a normal subgroup of a group $G$, and $M$ is a characteristic subgroup of $N$, then $M$ is a normal subgroup of G.

Prove that:

If $N$ is a normal subgroup of a group $G$, and $M$ is a characteristic subgroup of $N$, then $M$ is a normal subgroup of G.

Here what I am seeing is that $M$ is normal in $N$ and $N$ normal in $G$. But normality is not transitive property. So how to go?

• Please edit your post to include your thoughts and efforts on the problem: For example, what do you know about characteristic subgroups, and do you see any way in which you might use the definition? This will help people write responses that are appropriate to your question, and your question will likely be much better received. – user61527 Jan 22 '14 at 3:19
• Here what I am seeing is that $M$ is normal in $N$ and $N$ normal in $G$. But normality is not transitive property. So how to go? – user121418 Jan 22 '14 at 3:27
• @user121418, add that to the post.. – Apurv Jan 22 '14 at 3:30

Let $a \in G$ and $\varphi _a$ be the inner automorphism of $G$ induced by $a$. Then the restriction of $\varphi _a$ to $N$ is an automorphism of $N$ since $N$ is normal in $G$. Hence, $M$ is invariant under $\varphi _a$ for all $a ∈ G$, i.e., $M$ is normal in $G$.
Let f be an inner automorphism of G such that f : G $\to$ G . Then, the restriction of f to N is an automorphism of N since N is a normal subgroup of G, as previously stated. Here is where things get cleared up: Since M is a characteristic subgroup of N, it is invariant under f since all characteristic subgroups are invariant under all automorphisms. We can say this for our inner automorphism f since Inn(G) $\subset$ Aut(G). Since a subgroup that is invariant under all inner automorphisms is normal, M must be normal.
Short hand: M is characteristic $\Rightarrow$ M is invariant under all automorphisms $\Rightarrow$ M is normal. In our case, the automorphism we are talking about is f that holds from N to N since N is normal in G.