Proving Characteristic Subgroups are Transitive

Question

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?

Answer 1

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.

Answer 2

I understand your frustration, I also read several proofs but hard to understand them because somehow and I don't know why they all seem to omit the most important part of the explanation.

First about restriction notation, it's easy to understand and a neat thing write. So instead of writing:

$\phi(N) = N$ $\forall \phi \in Aut(K)$

It could be written as:

$\phi|_K(N) = N$

So we are going to use this notation to prove that if $N \: char \: K \: char \: G$ then $N \: char \: G$.

Let $\phi \in Aut(G)$, so $\phi(K) = K$ since $K$ is characteristic in $G$.

Now please note the most important part of the proof. If we have $\varphi \in Aut(K)$ we will also have $\varphi(K) = K$, so it's no difference than $\phi(K) = K$ as there is all the elements of $Aut(K)$ in $Aut(G)$. In other words:

$\varphi(K) = \phi(K) = K$.

Note that $\varphi$ could be written in restriction notation as $\phi|_K(K) = K$, it's the same thing because $\phi|_K \in Aut(K)$. So we could rewrite the above equality as:

$\phi|_K(K) = \phi(K) = K$.

( So we don't have to define a new symbol such $\varphi$ every time. )

As $N \: char \: K$, we have the same relationship between $N$ and $K$ as following:

$\phi|_N(N) = \phi|_K(N) = N$.

But we have had that $\phi|_K(K) = \phi(K)$, or is all there elements of $Aut(K)$ in $Aut(G)$, including $N < K$, so we have $\phi|_K(N) = \phi(N)$. Combine them all, we have:

$\phi|_N(N) = \phi|_K(N) = \phi(N) = N$.

Which is by definition $N \: char \: G$.