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Why is an image of a homomorphism a subgroup of the codomain? I know that the image of any function is a subset of the domain, but I don't know how to proceed with the former statement.

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    $\begingroup$ I guess the nontrivial part would be $1=\eta(1)=\eta(aa^{-1})=\eta(a)\eta(a^{-1})\implies \eta(a^{-1})=\eta(a)^{-1}$. $\endgroup$ – Pedro Tamaroff Sep 8 '13 at 2:52
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Prove that the propierties of a subgroup are valid for the elements of the image of a homomorphism. If $\varphi :G \to H$, you will have $e_H = \varphi(e_G)$, and $(\varphi(g))^{-1}=\varphi(g^{-1})$, for $g \in G$.

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