If $\phi :G\rightarrow H$ is a group homomorphism and $G$ is soluble, then $Im(\phi)$ is also soluble

Question

I'm trying to prove the following statement:

If $\phi :G\rightarrow H$ is a group homomorphism and $G$ is soluble, then $Im(\phi)$ is also soluble,

I tried creating a map $\psi:G\rightarrow Im(\phi)$, such that $\psi(g)=\phi(g)$, where $g\in G$. Clearly the map is surjective, but how can I show that it's injective. By doing so, I get that $G\cong Im(\phi)$. Would this be enough to show that $Im(\phi)$ is soluble.

Answer 1

It is enough to show that $\phi(G')=\phi(G)'$.

Observe that $\phi(xyx^{-1}y^{-1})=\phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1}$. Hence the result follows.

Edit: If $$\phi(G')=\phi(G)'$$ then $$\phi(G^r)=\phi(G)^r$$. By $G^r$ I mean $r$ th commutater subgroup of $G$.

Since $G$ is solvable, $G^n=1$ for some $n\implies 1=\phi(G^n)=\phi(G)^n$. Hence $\phi(G)$ is solvable.

Answer 2

HINTS

That $G$ is solvable means that it has a composition series with abelian quotients.

If you start with a composition series for $G$, how can you get one for $im(\phi)$?

Answer 3

$$\textrm{Ker}(\phi) \leq G$$

$$[G: \textrm{Ker}(\phi)] \simeq \textrm{Im}(\phi)$$

Quotient groups of solvable groups are solvable, so we are done