# Proving Characteristic Subgroups are Transitive

I know this is exactly the same as this question. But the proof detailed uses restrictions, and I'm not familiar with that. The way I want to prove this is by the standard method of showing two sets are equal since both $$\phi(G)$$ and $$G$$ are both sets correct?

If we have groups $$N,K,G$$ where $$N \leq K \leq G$$.
Given:
$$\phi(N) = N$$ $$\forall \phi \in Aut(K)$$
$$\phi(K) = K$$ $$\forall \phi \in Aut(G)$$
Show:
$$\phi(N) = N$$ $$\forall \phi \in Aut(G)$$

First we pick $$x \in \phi(N)$$ and show $$x \in N$$ where $$\phi: G \rightarrow G$$ is an automorphism.

However, this is where I'm not sure what to do. Am I even on the right track here?

• Maybe try looking up what restrictions are before dismissing them. I'm not sure there's a solution that doesn't use them. – jgon Apr 15 at 17:27
• I don't seem to think that saying "I'm not familiar with that" implies that he is dismissing them (i.e. treating them as unworthy of serious consideration) – Dean Young Apr 15 at 17:38

## 1 Answer

Maybe let's just talk through why restrictions is the way to go.

Suppose we have $$\phi\colon G \to G$$ an automorphism. We want to show that $$\phi(N) = N$$. Since $$K$$ is characteristic, we know that $$\phi(K) = K$$. Now consider the map $$\varphi\colon K \to K$$ defined by $$\varphi(k) = \phi(k)$$. Since $$\phi(K) = K$$, we see that $$\varphi$$ is surjective. I'll leave you to check that $$\varphi$$ is injective and a homomorphism. What we've shown then, is that $$\varphi$$ is an automorphism of $$K$$, so since $$N$$ is characteristic in $$K$$, we see that $$\varphi(N) = N$$. This tells us that $$\text{for every }x \in N,\ \varphi(x) = \phi(x) \in N.$$

Therefore we conclude that $$\phi(N) = N$$.

Actually, if you've done your exercise, you've shown something a little more: $$\varphi$$ is exactly what we mean by restriction: $$\phi|_K$$ means look at what happens if we forget about what $$\phi$$ does to group elements outside of $$K$$. Since $$K$$ is a subgroup, you've shown that $$\phi|_K$$ is still a homomorphism.