1. ## Characteristic normal subgroups

Prove that every characteristic subgroup of a normal subgroup of a group G is a normal subgroup of G, and that every characteristic subgroup of a characteristic subgroup of a group G is a characteristic subgroup of G.

Proof so far:

Suppose that N is a normal subgroup of G, and let H be a charcteristic subgroup of N. Find an inner automorphism of H such that $\displaystyle f:H \mapsto H by f(h) = nhn^{-1} \ \ \ n \in N$ with $\displaystyle f(H) = H$

Now, since N is normal, we have $\displaystyle gNg^{-1} \in N$, so we have $\displaystyle gHg^{-1} = g f(H) g^{-1} = g ( nhn^{-1} ) g^{-1} = (gn)h(gn)^{-1}$

So H is normal in G.

Is this right?

Suppose that N is a normal subgroup of G, and let H be a charcteristic subgroup of N. Find an inner automorphism of H such that $\displaystyle f:H \mapsto H by f(h) = nhn^{-1} \ \ \ n \in N$ with $\displaystyle f(H) = H$
Now, since N is normal, we have $\displaystyle gNg^{-1} \in N$, so we have $\displaystyle gHg^{-1} = g f(H) g^{-1} = g ( nhn^{-1} ) g^{-1} = (gn)h(gn)^{-1}$
What you wrote is messy. Let $\displaystyle H$ be a charachteristic subgroup. And let $\displaystyle g\in G$. We want to show $\displaystyle ghg^{-1} \in H$ for $\displaystyle h\in H$. Define $\displaystyle \theta: N\to N$ by $\displaystyle \theta (n) = gng^{-1}$ - this is an automorphism because $\displaystyle N$ is normal. Therefore, $\displaystyle \theta (H) = H$ and so $\displaystyle ghg^{-1} \in H$.