Let $G$ be a group; let $N$ be a normal subgroup of $G$; and let $M$ be a characteristic subgroup of $N$. Then how to show that $M$ is normal in $G$?

My work:

Let $m \in M$, $g \in G$. We need to show that $gmg^{-1} \in M$.

Now let $T$ be an automorphism of $G$ such that $g = T(g_0)$ for some $g_0 \in G$ and also $m = T(m_0)$ for some $m_0 \in G$.

We now show that $m_0$ is actually in $M$: Since $m = T(m_0)$, we can also write $m_0 = T^{-1}(m)$, where $T^{-1}$ denotes the inverse of the bijection $T$. And as $T^{-1}$ is also an automorphism, we can conclude that $m_0 \in M$.

Thus, $$ gmg^{-1} = T(g_0) T(m_0) T(g_0)^{-1} = T(g_0) T(m_0) T(g_0^{-1}) = T(g_0 m_0 g_0^{-1}).$$ Now as $m_0 \in M$ and $M \subset N$, we must have $m_0 \in N$; and since $N$ is a normal subgroup of $G$, it follows that $g_0 m_0 g_0^{-1} \in N$.

Also, we note that $$ gmg^{-1} = T(g_0) T(m_0) T(g_0)^{-1} = T_{T(g_0)}(T(m_0)) = (T_{T(g_0)} o T)(m_0), $$ and $T_{T(g_0)} o T$ is also an automorphism of $G$; thus it follows that $gmg^{-1} \in M$, showing that $M$ is normal in $G$.

Am I right? Or, is there any lacuna in my effort?

By a characteristic subgroup, we mean a subgroup that is invariant under all automorphisms of the parent group. That is, $M$ is a characteristic subgroup of $N$ if and only if $T[M] \subset M$ for all automorphisms $T$ of $N$. Then $M$ can be shown to be a normal subgroup of $N$.

Furthermore, for any $g \in G$, we define $T_g \colon G \to G$ as follows: $T_g(x) \colon = gxg^{-1} $ for all $x\in G$. This map can be shown to be an automorphism of $G$ and is called conjugation by $g$.

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    $\begingroup$ You are assuming that $M$ is a characteristic subgroup of $N$. But in the beginning it seems that you are using $T(M) \subseteq M$ for any automorphism $T$ of $G$, which might not necessarily hold (since $M$ might not be characteristic in $G$). Here is a hint: $x \mapsto gxg^{-1}$ defines an automorphism $N \rightarrow N$ $\endgroup$ – spin May 18 '14 at 11:13
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    $\begingroup$ There is no reason to expect any such $T$ to exist. Instead, think about the expression $gmg^{-1}$. How does that generalize when you replace $m$ by an arbitrary element of $N$? $\endgroup$ – Lee Mosher May 18 '14 at 12:36

If $g \in G $, then $\phi_g$ defined as $$\phi_g(h) = g^{-1} \cdot h \cdot g$$ is an automorphism of $G$.

Note that $\phi_g$ restricted to $N$ is an automorphism of $N$ because $N$ is normal in $G$, thus $$\phi_g(M) \subseteq \phi_g(N) = N$$ but $M$ is characteristic in $N$ and so $$M^g = \phi_g(M) = M \ \ \ \forall g \in G$$ This means that $M$ is normal in $G$


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