The image of a group homomorphism is a subgroup

Let $𝐺$ and $𝐻$ be groups, and $𝑓 : 𝐺 \to 𝐻$ be a Group homomorphism. Then the image

𝑓 (𝐺) = {𝑓 (𝑔) : 𝑔 \in 𝐺}

is a subgroup of $𝐻$ . #m/thm/group

Proof

Since $𝑓 (𝑒) = 𝑒$ , clearly $𝑒 \in 𝑓 (𝐺)$ . Let $𝑎, 𝑏 \in 𝑓 (𝐺)$ . Then there exist (not necessarily unique) $𝑥, 𝑦 \in 𝐺$ such that $𝑓 (𝑥) = 𝑎$ and $𝑓 (𝑦) = 𝐺$ . It follows that $𝑎 𝑏 - 1 = 𝑓 (𝑥) 𝑓 (𝑦) - 1 = 𝑓 (𝑥 𝑦 - 1) \in 𝑓 (𝐺)$ . Therefore $𝑓 (𝐺)$ is a subgroup by One step subgroup test.

Corollary

It follows that the image of a subgroup is also a subgroup, since a group homomorphism induces a subgroup homomorphism.

#state/tidy | #lang/en | #SemBr