Exercise 1.6.14 (Kernels of homomorphisms)

Proof. By definition , $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\subset G$.

• Since $\phi (1_G)=1_H$, we know that $1_G\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, so $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\ne \varnothing$.
• If $g,g^{\prime }\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $\phi (g{g}^{\prime })=\phi (g)\phi (g^{\prime })=1_G$, thus $g{g}^{\prime }\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$.
• If $g\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $\phi (g^{-1})=\phi {(g)}^{-1}=1_G^{-1}=1_G$, so $g^{-1}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$

Since $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\ne \varnothing$ is stable for product and inverses, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ is a subgroup of $G$.

Since $1_G\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, $\{1_G\}\subset \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$.

Suppose that $\phi$ is injective. If $g\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $\phi (g)=1_H=\phi (1_G)$. Since $\phi$ is injective, $g=1_G$, thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\subset \{1_G\}$. This shows that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1_G\}$.

Conversely, suppose that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1_G\}$. If $g,g^{\prime }\in G$ are such that $\phi (g)=\phi (g^{\prime })$, then $\phi (g{g{}^{\prime }}^{-1})=\phi (g)\phi {(g^{\prime })}^{-1}=1_H$, so $g{g{}^{\prime }}^{-1}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$. The hypothesis $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1_G\}$ shows that $g{g{}^{\prime }}^{-1}=1_G$, thus $g=g^{\prime }$. This proves that $\phi$ is injective.

In conclusion, $\phi$ is injective if and only if the kernel of $\phi$ is the identity subgroup of $G$. □