2.4.3

For kernels we have $$x, y \in \text{ker}\varphi \implies \varphi(xy) = \varphi(x)\varphi(y) = 1 \cdot 1 = 1 \implies xy \in \text{ker}\varphi$$; and $$\varphi(1) = 1 \implies 1 \in \text{ker}\varphi$$; and $$x \in \text{ker}\varphi \implies \varphi(x^{-1}) = \varphi(x)^{-1} = 1^{-1} = 1 \implies x^{-1} \in \text{ker}\varphi.$$

For images we have $$x, y \in \text{im} \varphi \implies$$ there exists $$z_0$$, $$z_1$$ s.t. $$xy = \varphi(z_0)\varphi(z_1) = \varphi(z_0 z_1)$$ and as $$G$$ is closed under composition $$z_0 z_1 \in G \implies xy \in \text{im} \varphi$$; and $$\varphi(1) = 1 \implies 1 \in \text{im}\varphi$$; and $$x \in \text{im}\varphi \implies$$ there exists $$z_0$$ s.t. $$x = \varphi(z_0) \implies x^{-1}x\varphi(z_0)^{-1} = x^{-1}\varphi(z_0)\varphi(z_0)^{-1} \implies \varphi(z_0)^{-1} = x^{-1} $$$$ \implies \phi(z_0^{-1}) = x^{-1}$$ and as $$z_0^{-1} \in G$$ as $$G$$ is a group $$x^{-1} \in \text{im} \varphi.$$