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

Definition

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. (Proof.)

• The inverse of the image is the image of the inverse. (Proof.)

• The image of a group under a homomorphism is another group. (Proof.)

• The composition of two homomorphisms is a homomorphism. (Proof.)