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  text: 'Let $f: G \\to H$ be a [-47t], and write $f(G)$ for the set $\\{ f(g) : g \\in G \\}$.\nThen $f(G)$ is a group under the operation inherited from $H$.\n\n# Proof\n\nTo prove this, we must verify the group axioms.\nLet $f: G \\to H$ be a group homomorphism, and let $e_G, e_H$ be the identities of $G$ and of $H$ respectively.\nWrite $f(G)$ for the image of $G$.\n\nThen $f(G)$ 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 $f(G)$.\n\n$e_H$ 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.\n\nInverses exist, by "the inverse of the image is the image of the inverse".\n\nThe operation remains associative: this is inherited from $H$.\n\nTherefore, $f(G)$ is a group, and indeed is a subgroup of $H$.',
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