[summary: A group homomorphism is a function between groups which "respects the group structure".]

[summary(Technical): Formally, given two groups $~$(G, +)$~$ and $~$(H, *)$~$ (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(a) * f(b) = f(a+b)$~$ for all $~$a, b \in G$~$.]

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

# Definition

Formally, given two groups $~$(G, +)$~$ and $~$(H, *)$~$ (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(a) * f(b) = 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 $~$1_G(g) = g$~$ for all $~$g \in G$~$. This homomorphism is always bijective.
- For any group $~$G$~$, there is a (unique) group homomorphism into the group $~$\{ e \}$~$ with one element and the only possible group operation $~$e * 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^{\mathrm{op}}$~$ given by taking inverses: $~$g \mapsto g^{-1}$~$. The group $~$G^{\mathrm{op}}$~$ is defined to have underlying set equal to that of $~$G$~$, and group operation $~$g +_{\mathrm{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(e_{C_2}) = f(gg) = f(g) f(g) = h h = h^2$~$, which is not true. (We have used that the identity gets mapped to the identity.)

## Comments

Patrick Stevens

I have a question about general Arbital practice here. A mathematician will probably already know what a group homomorphism is, but they probably also don't need the proofs of the Properties, for instance, and they don't need the explanation of the trivial group. Should I have split this up into different lenses in some way?