Proof of Kernel and Image of a Group Homomorphism are Subgroups

We write both operations multiplicatively and denote by $e_G$ and $e_H$ the identity elements of $G$ and $H$, which exist and are unique by Uniqueness of the Identity Element and of Inverses in a Group. Throughout we use the homomorphism property $\varphi(ab)=\varphi(a)\varphi(b)$ from Group Homomorphism and Isomorphism, and the identities $\varphi(e_G)=e_H$ and $\varphi(a^{-1})=\varphi(a)^{-1}$ from Group Homomorphisms Preserve the Identity Element and Inverses.

The kernel is a subgroup of $G$. We verify the three conditions of Subgroup for the kernel of $\varphi$.

Condition 1. Since $\varphi(e_G)=e_H$, we have $e_G\in\ker\varphi$.

Condition 2. Let $a,b\in\ker\varphi$, so $\varphi(a)=e_H$ and $\varphi(b)=e_H$. Then

using condition 2 of Group and Abelian Group in $H$ for the last equality. Hence $ab\in\ker\varphi$.

Condition 3. Let $a\in\ker\varphi$. Then

where the last equality is claim 4 of Cancellation Laws and Basic Inverse Identities in a Group applied in $H$. Hence $a^{-1}\in\ker\varphi$.

The image is a subgroup of $H$. We verify the three conditions of Subgroup for the image of $\varphi$, this time as a subset of $H$.

Condition 1. Since $\varphi(e_G)=e_H$, the element $e_H$ lies in $\operatorname{im}\varphi$.

Condition 2. Let $y,z\in\operatorname{im}\varphi$, say $y=\varphi(a)$ and $z=\varphi(b)$ with $a,b\in G$. Then $ab\in G$ and

so $yz\in\operatorname{im}\varphi$.

Condition 3. Let $y\in\operatorname{im}\varphi$, say $y=\varphi(a)$ with $a\in G$. Then $a^{-1}\in G$ and

so $y^{-1}\in\operatorname{im}\varphi$.