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Let $\phi : G \rightarrow G'$ be a homomorphism. We want to show that $H \leq G$ implies that $\phi(H) \leq G'$.

This is one of those problems that seems obviously true to the intuition, yet I'm having a devil of a time trying to prove it formally. A sketch of my efforts so far:

Let us define the codomain of $H$ under $\phi$ to be $H'$, and $h'$ to be an arbitrary element of $H'$ such that $\phi(h)=h'$ (for $h \in H$).

To achieve the desired result, we have to show that $h' \in G'$.

What I want to say is that, since $H \leq G$, then every $h$ in $H$ is also in $G$, and thus clearly $h' \in G'$. Is it that simple, or am I begging the question?

JustSomeGuy716
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  • If $a, b \in H$ then $\phi(ab^{-1}) \in H'.$ Now use the fact that you have a homomorphism. – Sean Roberson Mar 16 '19 at 17:47
  • You have shown that $H’$ is a $\textit{subset}$ of $G’$, not that it is a $\textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.) – mindfields Mar 16 '19 at 17:52

1 Answers1

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Hint: You have to show that, for $h_1,h_2$ we have $\phi(h_1)\phi(h_2)\in G'$ and $\phi(h_1)^{-1}\in G'$. Then $\phi(H)$ is a subgroup of $G'$.

Reference: This duplicate:

Image of subgroup and Kernel of homomorphism form subgroups

Dietrich Burde
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