**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.14**

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 $\mathsf{ker}\ \varphi = 1$.

Solution: By Exercise 1.1.26, 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.