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 \(H\).


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 \(H\).


  • Group homomorphism

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