Showing the image of a subgroup is a subgroup.

Question

Let $G$ and $G'$ be groups. Let $\phi: G \to G'$ be a homomorphism. Let $H$ be a group of $G$ and let $H'=\{\phi(h): h\in H\}$. Want to show that $H'$ is a subgroup.

Proof:

(i) $H'$ has identity since $\phi(e_G)\in H'$. This is because $\phi(e_G)=e_{G'}$.

(ii) $H'$ is closed because of the following: Let $h_{1}^{'}, h_{2}^{'} \in H'$. Then there exists an $h_1,h_2$ in $H$ such that $h_1^{'}=\phi(h_1)$ and $h_2=\phi(h_{2}^{'})$. Then $h_1^{'}h_2^{'}=\phi(h_1)\phi(h_2)=\phi(h_1h_2)$.

(iii) For each $h^{'} \in H'$ there exists an $h \in H$ such that $\phi(h)=h'$. But, $h'^{-1}=\phi(h)^{-1}=\phi(h^{-1})$. Hence it has inverses.

Did I prove this correctly?

You have put the prime on the wrong $h_2$ in the proof of (ii) [$h_2 = \phi(h_2')$ instead of $h_2' = \phi(h_2)$], but apart from that typo, it's correct. — Daniel Fischer, Oct 01 '13 at 20:02
I'll admit, I got a little to trigger happy with the primes. — Vikram, Oct 01 '13 at 20:03

score 1 · Answer 1 · answered Oct 01 '13 at 20:18

Looks correct. I'd rephrase (i) as "$H'$ has identity since $e_{G'} = \phi(e_G)\in H'$." With your formulation "$H'$ has identity since $\phi(e_G)\in H'$. This is because $\phi(e_G)=e_{G'}$" it looks like you're saying that $\phi(e_G)=e_{G'}$ implies $\phi(e_G)\in H'$, but that's not what you mean to say.

Showing the image of a subgroup is a subgroup.

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