# Group homomorphism

https://arbital.com/p/group_homomorphism

by Patrick Stevens Jun 13 2016 updated Jun 22 2016

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

[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.)

# 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.)