Theorem. The homomorphic image of a group is a group. More detailed, if $f$ is a homomorphism from the group $(G,\,\ast)$ to the groupoid $(\Gamma,\,\star)$ , then the groupoid $(f(G),\,\star)$ also is a group. Especially, the isomorphic image of a group is a group.

Proof. Let $\alpha,\,\beta,\,\gamma$ be arbitrary elements of the image $f(G)$ and $a,\,b,\,c$ some elements of $G$ such that $f(a) = \alpha,\, f(b) = \beta,\, f(c) = \gamma$ . Then $$\alpha \star \beta \;=\; f(a) \star f(b) \;=\; f(a \ast b) \;\in\; f(G),$$ whence $f(G)$ is closed under ``$\star$ '', and we, in fact, can speak of a groupoid $(f(G),\,\star)$ .

Secondly, we can calculate

Let $e$ be the identity element of $(G,\,\ast)$ and $f(e) = \varepsilon$ . Then $$\varepsilon \star \alpha \;=\; f(e) \star f(a) \;=\; f(e \ast a) \;=\; f(a) \;=\; \alpha,$$ $$\alpha \star \varepsilon \;=\; f(a) \star f(e) \;=\; f(a \ast e) \;=\; f(a) \;=\; \alpha,$$ and therefore $\varepsilon$ is an identity element in $f(G)$ .

If $f(a^{-1}) = \alpha'$ , then $$\alpha \star \alpha' \;=\; f(a) \star f(a^{-1}) \;=\; f(a \ast a^{-1}) \;=\; f(e) \;=\; \varepsilon,$$ $$\alpha' \star \alpha \;=\; f(a^{-1}) \star f(a) \;=\; f(a^{-1} \ast a) \;=\; f(e) \;=\; \varepsilon.$$ Thus any element $\alpha$ of $f(G)$ has in $f(G)$ an inverse.

Accordingly, $(f(G),\, \star)$ is a group.

Remark 1. If $(G,\,\ast)$ is Abelian, the same is true for $(f(G),\, \star)$ .

Remark 2. Analogically, one may prove that the homomorphic image of a ring is a ring.

Example. If we define the mapping $f$ from the group $(\mathbb{Z},\,+)$ to the groupoid $(\mathbb{Z}_9,\,\cdot)$ by $$f(n) \;:=\; \langle4\rangle^n,$$ then $f$ is homomorphism: $$f(m\!+\!n) \;=\; \langle4\rangle^{m+n} \;=\; \langle4\rangle^m\langle4\rangle^n \;=\; f(m)f(n).$$ The image $f(\mathbb{Z})$ consists of powers of the residue class $\langle4\rangle$ , which are $$\langle4\rangle,\;\; \langle16\rangle = \langle7\rangle,\;\; \langle64\rangle = \langle1\rangle.$$ These apparently form the cyclic group of order 3.