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For a finite group $G$, define $ R(G)=\cap\{K \triangleleft G \; | \; G/K \text{ is solvable}\}$ If $\alpha:G\rightarrow G_1$ is a group homomorphism, show that $\alpha[R(G)]\subseteq R(G_1)$.

This is part $(c)$ of the question - I've already proved that $R$ is the smallest normal subgroup of $G$ such that $G/R$ is solvable, so maybe that would come in useful. Thanks in advance!

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Remember also that the epimorphic image of a normal subgroup is again normal...

$x\in R(G)\Longrightarrow \forall\,K\triangleleft G\,\,s.t.\,\,G/K\,\,\text{solvable}\,\,,\,x\in K\Longrightarrow \alpha(x)\in\alpha(K)$

But (assuming $\,\alpha\,$ is a epimorphism) ,we have that $\,G_1\cong G/\ker\alpha\,$ , so in fact

$\alpha(K)=K\ker\alpha/\ker\alpha\cong K/\left(K\cap\ker\alpha\right)\Longrightarrow$

$\Longrightarrow G_1/\alpha(K)=\left(G/\ker\alpha\right)/\left(K\ker\alpha/\ker\alpha\right)\cong G/K\ker\alpha\cong\left(G/K\right)/\left(K\ker\alpha/K\right)$

and since the last group on the right is a quotient group of the solvable group $\,G/K\,$ , so is the group $\,G_1/\alpha(K)\,$ solvable.

From here, $\,\alpha(x)\in R(G_1)\,$ and we're done.

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    Got it. Thank you both!!2012-11-22