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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.

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    $\begingroup$ An arbitrary group homomorphism is clearly not injective. $\endgroup$
    – anomaly
    Mar 24 '15 at 19:36
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    $\begingroup$ Have you heard about quotient group and first isomorphism theorem? $\endgroup$
    – Berci
    Mar 24 '15 at 19:37
  • $\begingroup$ yes, but why would that imply that the group homomorphism isn't injective. I thought that if $\phi(g)=\phi(h)$, then $\phi(g)-\phi(h)=\phi(g-h)=0$. And so, $g-h=0$ which implies $g=h$ $\endgroup$ Mar 24 '15 at 19:39
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    $\begingroup$ By the way, even if soluble is sometimes used, one generally rathers says solvable. $\endgroup$
    – Olórin
    Mar 24 '15 at 19:39
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    $\begingroup$ Using additive notation isn't helpful, but in any case, if we have $\phi(g) = \phi(h)$ then all we can conclude is that $gh^{-1} \in {\rm ker} \phi$. It need not be the case that ${\rm ker} \phi = \{1_{G} \}.$ $\endgroup$ Mar 24 '15 at 19:43
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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.

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  • $\begingroup$ Sorry I'm not sure what you mean by $G'$ $\endgroup$ Mar 24 '15 at 20:03
  • $\begingroup$ @AndrewBrick: The Commutator subgroup. Check this. Generally defination of solvable groups is done by derived series. en.wikipedia.org/wiki/Commutator_subgroup $\endgroup$
    – mesel
    Mar 24 '15 at 20:05
  • $\begingroup$ So if $\phi(G')=\phi(G)'$, then how does this show that $Im(\phi)$ is soluble. Sorry if the question seems obvious $\endgroup$ Mar 24 '15 at 20:18
  • $\begingroup$ @AndrewBrick: Ok. I will edit my answer by giving details. $\endgroup$
    – mesel
    Mar 24 '15 at 20:23
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    $\begingroup$ @AndrewBrick: $G'=<\{xyx^{-1}y^{-1}|x,y\in G\}>$ Then $G'$ is called as Commutator subgroup of $G$. $G'$ has many property; $G/G'$ must be an abelian group for example. $G^2=G''$ i.e Commutator of the commutator groups. Hence, $G^r=G'''''''...$. Notice that $G^r/G^{r+1}$ is abelian. Hence $G\geq G'\geq ... G^r\geq ..$ is an composition series. Hence when $G$ is solvable $G^n$ reach the identity. $\endgroup$
    – mesel
    Mar 24 '15 at 20:46
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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 $im(\phi)$?

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  • $\begingroup$ So if $1\subset G_1\subset...\subset G_n=G$ is a composition series for $G$, then how can I get one for $Im(\phi)$ $\endgroup$ Mar 24 '15 at 19:58
  • $\begingroup$ @AndrewBrick Try applying $\phi$ to each part of the series. $\endgroup$ Mar 25 '15 at 7:52
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$$\textrm{Ker}(\phi) \leq G$$

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

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

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