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 $N \leq K \leq G$.
$\phi(N) = N$ $\forall \phi \in Aut(K)$
$\phi(K) = K$ $\forall \phi \in Aut(G)$
$\phi(N) = N$ $\forall \phi \in Aut(G)$

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?

  • $\begingroup$ Maybe try looking up what restrictions are before dismissing them. I'm not sure there's a solution that doesn't use them. $\endgroup$ – jgon Apr 15 at 17:27
  • $\begingroup$ 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) $\endgroup$ – Dean Young Apr 15 at 17:38

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.


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