[SOLVED] Homomorphisms groups and subgroup help

**kpizle** · November 8th 2009, 07:25 PM

Been working for hours on my homework...brain is fried. Can not figure out this last problem. Any help/guidance would be appreciated. Thanks.

Prove or disprove that if $phi$ and $psi$ are homomorphisms between groups G and G', then the set

H = { x element of G | $psi$ (x) = $phi$ (x)}

is a subgroup of G.

thanks again!

**Drexel28** · November 8th 2009, 09:55 PM

Originally Posted by kpizle

Been working for hours on my homework...brain is fried. Can not figure out this last problem. Any help/guidance would be appreciated. Thanks.

Prove or disprove that if $phi$ and $psi$ are homomorphisms between groups G and G', then the set

H = { x element of G | $psi$ (x) = $phi$ (x)}

is a subgroup of G.

thanks again

Problem: Let $\Phi:G\mapsto G'$ and $\Psi:G\mapsto G'$ be homomorpisms. Is the set $\mathcal{H}=\left\{g\in G:\Phi(g)=\Psi(g)\right\}$ a subgroup of $G$ ? Prove or disprove your answer.

Claim: It is.

Proof: Let us verify the four neccessary conditions to be a subgroup

1. Clearly $\mathcal{H}$ will inherit $G$ 's associativity.

2. You should know that $e\in\text{Ker}\left(\Phi\right),\text{Ker}\left(\P si\right)$ . Thus $\Phi\left(e_G\right)=e_{G'}=\Psi\left(e_G\right)\i mplies e_G\in\mathcal{H}$ .

3. Now suppose that $g\in\mathcal{H}$ , then $\Phi\left(g\right)=\Psi\left(g\right)$ , so that $\Phi^{-1}(g)\Psi(g)=\Phi^{-1}(g)\Psi(g)=e_{G'}\implies \Phi^{-1}(g)=\Psi\left(g^{-1}\right)\implies\Phi\left(g^{-1}\right)=\Psi\left(g^{-1}\right)$ which of course implies $g^{-1}\in\mathcal{H}$ .

4. Suppose $g,g'\in\mathcal{H}$ then $\Phi\left(g\right)=\Psi\left(g\right)$ and $\Phi\left(g'\right)=\Psi\left(g'\right)$ . Therefore $\Phi\left(gg'\right)=\Phi(g)\Phi\left(g'\right)=\P si\left(g\right)\Psi\left(g'\right)=\Psi\left(gg'\ right)$ . Therefore $gg'\in\mathcal{H}$

This completes the proof.

Remark: I always find the direct satsifaction of the subgroup axioms to be more instructive. You may (and should) attempt to redo this using the condition that if $a,b\in\mathcal{H}\implies ab^{-1}\in\mathcal{H}$ .

**kpizle** · November 9th 2009, 05:17 AM

Drexel28:

Thank you very much.

After a night's sleep, this seems much clearer.

I was not sure whether to try to prove with the 3 conditions of a subgroup, or the Subgroup Criterion. After attempting both (after working 8 other of these suckers), I felt like I was doing it wrong. But looking at my attempt and your reply, I was on the right track.

Thanks again!

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