The image of a group homomorphism is a subgroup

Let G and H be groups and let \varphi : G \rightarrow H be a group homomorphism. Prove that the image \mathsf{im}\ \varphi of \varphi is a subgroup of H. Prove that if \varphi is injective then G \cong \mathsf{im}\ \varphi.

By a previous exercise, it suffices to show that \mathsf{im}\ \varphi is closed under multiplication and inversion. To that end, let h_1, h_2 \in \mathsf{im}\ \varphi. Then there exist g_1, g_2 \in G such that \varphi(g_1) = h_1 and \varphi(g_2) = h_2. Since h_1 h_2 = \varphi(g_1) \varphi(g_2) = \varphi(g_1 g_2), \mathsf{im}\ \varphi is closed under multiplication. Now let h \in \mathsf{im}\ \varphi; then there exists g \in G such that \varphi(g) = h. Now h^{-1} = \varphi(g)^{-1} = \varphi(g^{-1}), so that \mathsf{im}\ \varphi is closed under inversion. Thus \mathsf{im}\ \varphi is a subgroup of H.

Suppose now that \varphi is injective. We can then defined a mapping \psi : G \rightarrow \mathsf{im}\ \varphi by \psi(g) = \varphi(g); this \psi is clearly a homomorphism and is bijective, hence an isomorphism. Thus G \cong \mathsf{im}\ \varphi.

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