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?
 A: 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 $G$.
((The Theory of Finite Groups An Introduction By Hans Kurzweil ,Bernd Stellmacher.page 17))
A: 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.   
A: Since M is characteristic in N, any automorphism of N will send M back to itself. But since N is normal in G, conjugation by any g in G is an automorphism of N.
