A group homomorphism is a function between groups which "respects the group structure".

Formally, given two groups $(G,+)$ and $(H,\ast )$ (which hereafter we will abbreviate as $G$ and $H$ respectively), a group homomorphism from $G$ to $H$ is a function $f$ from the underlying set $G$ to the underlying set $H$, such that $f\left(a\right)\ast f\left(b\right)=f(a+b)$ for all $a,b\in G$.

Examples

• For any group $G$, there is a group homomorphism $1_G:G\to G$, given by $1_G\left(g\right)=g$ for all $g\in G$. This homomorphism is always bijective.
• For any group $G$, there is a (unique) group homomorphism into the group $\left\{e\right\}$ with one element and the only possible group operation $e\ast e=e$. This homomorphism is given by $g\mapsto e$ for all $g\in G$. This homomorphism is usually not injective: it is injective if and only if $G$ is the group with one element. (Uniqueness is guaranteed because there is only one function, let alone group homomorphism, from any set $X$ to a set with one element.)
• For any group $G$, there is a (unique) group homomorphism from the group with one element into $G$, given by $e\mapsto e_G$, the identity of $G$. This homomorphism is usually not surjective: it is surjective if and only if $G$ is the group with one element. (Uniqueness is guaranteed this time by the property proved below that the identity gets mapped to the identity.)
• For any group $(G,+)$, there is a bijective group homomorphism to another group $G^{op}$ given by taking inverses: $g\mapsto g^{-1}$. The group $G^{op}$ is defined to have underlying set equal to that of $G$, and group operation $g{+}_{op}h:=h+g$.
• For any pair of groups $G,H$, there is a homomorphism between $G$ and $H$ given by $g\mapsto e_H$.
• There is only one homomorphism between the group $C_2=\{e_{C_2},g\}$ with two elements and the group $C_3=\{e_{C_3},h,h^2\}$ with three elements; it is given by $e_{C_2}\mapsto e_{C_3},g\mapsto e_{C_3}$. For example, the function $f:C_2\to C_3$ given by $e_{C_2}\mapsto e_{C_3},g\mapsto h$ is not a group homomorphism, because if it were, then $e_{C_3}=f\left(e_{C_2}\right)=f\left(gg\right)=f\left(g\right)f\left(g\right)=hh=h^2$, which is not true. (We have used that the identity gets mapped to the identity.)

Properties

The identity gets mapped to the identity

For any group homomorphism $f:G\to H$, we have $f\left(e_G\right)=e_H$ where $e_G$ is the identity of $G$ and $e_H$ the identity of $H$.

Indeed, $f\left(e_G\right)f\left(e_G\right)=f\left(e_G{e}_G\right)=f\left(e_G\right)$, so premultiplying by $f(e_G{)}^{-1}$ we obtain $f\left(e_G\right)=e_H$.

The inverse of the image is the image of the inverse

For any group homomorphism $f:G\to H$, we have $f\left(g^{-1}\right)=f(g{)}^{-1}$.

Indeed, $f\left(g^{-1}\right)f\left(g\right)=f\left(g^{-1}g\right)=f\left(e_G\right)=e_H$, and similarly for multiplication on the left.

The image of a group under a homomorphism is another group.

To prove this, we must verify the group axioms. Let $f:G\to H$ be a group homomorphism, and let $e_G,e_H$ be the identities of $G$ and of $H$ respectively. Write $f\left(G\right)$ for the image of $G$.

Then $f\left(G\right)$ is closed under the operation of $H$: since $f\left(g\right)f\left(h\right)=f\left(gh\right)$, so the result of $H$-multiplying two elements of $f\left(G\right)$ is also in $f\left(G\right)$.

$e_H$ is the identity for $f\left(G\right)$: it is $f\left(e_G\right)$, so it does lie in the image, while it acts as the identity because $f\left(e_G\right)f\left(g\right)=f\left(e_{G}g\right)=f\left(g\right)$, and likewise for multiplication on the right.

Inverses exist, by "the inverse of the image is the image of the inverse".

The operation remains associative: this is inherited from $H$.

Therefore, $f\left(G\right)$ is a group, and indeed is a subgroup of $H$.

The composition of two homomorphisms is a homomorphism

To prove this, note that $g\left(f\right(x\left)\right)g\left(f\right(y\left)\right)=g\left(f\right(x\left)f\right(y\left)\right)$ since $g$ is a homomorphism; that is $g\left(f\right(xy\left)\right)$ because $f$ is a homomorphism.