# Intersection of solvable quotient groups under homomorphism

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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What do you mean by $K \text{ker} \alpha / \text{ker} \alpha$? I'm a little confused by the notation here. – Samuel Reid Nov 22 '12 at 4:55
Put $\,N:=\ker\alpha\,$ , then the homomorphic image of a subgroup $\,K\leq G\,$ in the quotient $\,G/N\,$ is $\,KN/N\,$ – DonAntonio Nov 22 '12 at 5:01
What do you mean "in the quotient $G/N$?" Isn't it the fact that $\alpha(K) \cong K/ ker\alpha$? I don't quite understand where the other $ker\alpha$ on the top came from. – Samuel Reid Nov 22 '12 at 5:10
@Samuel To get the image of $K$ under $\alpha$, you can't just take $K/N$ because $N$ may not be a subgroup of $K$. Instead you have to form the product subgroup $KN$ first (which definitely contains $N$ as a subgroup), then take the quotient: $KN/N$ – Ted Nov 22 '12 at 6:08
Got it. Thank you both!! – Samuel Reid Nov 22 '12 at 8:23