1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.