# Characteristic normal subgroups

• Sep 25th 2008, 06:52 AM
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 $f:H \mapsto H by f(h) = nhn^{-1} \ \ \ n \in N$ with $f(H) = H$

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

So H is normal in G.

Is this right?
• Sep 25th 2008, 07:58 AM
ThePerfectHacker
Quote:

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 $f:H \mapsto H by f(h) = nhn^{-1} \ \ \ n \in N$ with $f(H) = H$

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

So H is normal in G.

Is this right?

What you wrote is messy. Let $H$ be a charachteristic subgroup. And let $g\in G$. We want to show $ghg^{-1} \in H$ for $h\in H$. Define $\theta: N\to N$ by $\theta (n) = gng^{-1}$ - this is an automorphism because $N$ is normal. Therefore, $\theta (H) = H$ and so $ghg^{-1} \in H$.