# [SOLVED] Homomorphisms groups and subgroup help

• Nov 8th 2009, 07:25 PM
kpizle
[SOLVED] Homomorphisms groups and subgroup help
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 $\displaystyle phi$ and $\displaystyle psi$ are homomorphisms between groups G and G', then the set

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

is a subgroup of G.

thanks again!
• Nov 8th 2009, 09:55 PM
Drexel28
Quote:

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 $\displaystyle phi$ and $\displaystyle psi$ are homomorphisms between groups G and G', then the set

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

is a subgroup of G.

thanks again

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

Claim: It is.

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

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

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

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

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

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 $\displaystyle a,b\in\mathcal{H}\implies ab^{-1}\in\mathcal{H}$.
• Nov 9th 2009, 05:17 AM
kpizle
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!