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 \(H\).