# Conditions under which group solvability is invariant under an homomorphism.

Let $G$ be a solvable group and $f: G \rightarrow H$ an isomorphism. My question is if $H$ is a solvable group.

Let $1=G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_n = G$ be a normal series of $G$ whose factors are abelian groups. Clearly, $1=f(G_0) \trianglelefteq f(G_1) \trianglelefteq ... \trianglelefteq f(G_n) = H$ is a normal series of $H$ but are the quotients abelian?

Is there a weaker condition for invariance?

If $N\triangleleft G$ is a normal subgroup and $f:G\to H$ is any group homomorphism, then $f(N)\triangleleft f(G)$ and there is an onto hommomorphism $G/N\to f(G)/f(N)$ given by $gN\mapsto f(g)f(N)$. The homomorphic image of any abelian group is abelian. Now apply to all composition factors.
Note this applies to showing any homomorphic image of a solvable group is solvable. The case when $f$ is an isomorphism out of $G$ should intuitively be trivial, since any two isomorphic groups have the exact same group-theoretic properties.
• i thought that the image of a normal subgroup is a normal subgroup provided that $f$ is an epimorphism not just an homomorphism...am i right? – Rodrigo Aug 12 '16 at 16:33
• @Rodrigo $G\to f(G)$ is an epimorphism, so $f(N)$ is normal in $f(G)$. – arctic tern Aug 12 '16 at 16:33