# The image of a group under a homomorphism is a subgroup of the codomain

Let $$f: G \to H$$ be a group homomorphism, and write $$f(G)$$ for the set $$\{ f(g) : g \in G \}$$. Then $$f(G)$$ is a group under the operation inherited from $$H$$.

# Proof

To prove this, we must verify the group axioms. Let $$f: G \to H$$ be a group homomorphism, and let $$e_G, e_H$$ be the identities of $$G$$ and of $$H$$ respectively. Write $$f(G)$$ for the image of $$G$$.

Then $$f(G)$$ is closed under the operation of $$H$$: since $$f(g) f(h) = f(gh)$$, so the result of $$H$$-multiplying two elements of $$f(G)$$ is also in $$f(G)$$.

$$e_H$$ is the identity for $$f(G)$$: it is $$f(e_G)$$, so it does lie in the image, while it acts as the identity because $$f(e_G) f(g) = f(e_G g) = f(g)$$, and likewise for multiplication on the right.

Inverses exist, by “the inverse of the image is the image of the inverse”.

The operation remains associative: this is inherited from $$H$$.

Therefore, $$f(G)$$ is a group, and indeed is a subgroup of $$H$$.

Parents:

• Group homomorphism

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