The kernel of a group homomorphism is a subgroup

Let \varphi : G \rightarrow H be a group homomorphism. Define the kernel of \varphi to be \mathsf{ker}\ \varphi = \{ g \in G \ |\ \varphi(g) = 1 \}. Prove that \mathsf{ker}\ \varphi is a subgroup of G. Prove that \varphi is injective if and only if \mathsf{ker}\ \varphi = 1.


By a previous example, it suffices to show that \mathsf{ker}\ \varphi is closed under multiplication and inversion. To that end, let g_1, g_2 \in \mathsf{ker}\ \varphi. Then \varphi(g_1 g_2) = \varphi(g_1) \varphi(g_2) = 1 \cdot 1 = 1, so that g_1g_2 \in \mathsf{ker}\ \varphi. Now let g \in \mathsf{ker}\ \varphi. Then \varphi(g^{-1}) = \varphi(g)^{-1} = 1^{-1} = 1 so that g^{-1} \in \mathsf{ker}\ \varphi. Thus \mathsf{ker}\ \varphi is a subgroup of G.

Now we prove the second statement.

(\Rightarrow) Suppose \varphi is injective, and let g \in \mathsf{ker}\ \varphi. Note that \varphi(1) = 1 = \varphi(g), so that g = 1. Thus \mathsf{ker}\ \varphi = 1.

(\Leftarrow) Suppose \mathsf{ker}\ \varphi = 1, and let a,b \in G such that \varphi(a) = \varphi(b). Now 1 = \varphi(a) \varphi(b)^{-1} = \varphi(a) \varphi(b^{-1}) = \varphi(ab^{-1}), so that 1 = ab^{-1} and thus a = b. Hence \varphi is injective.

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