Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.