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  "body": "\n## Proof\n\n1. Let $\\varphi \\colon G \\to H$ be a [[group homomorphism|0HV9Y7WG5R]], and let $e_G$ and $e_H$ be the identity elements of $G$ and $H$.\n2. Since $e_G e_G = e_G$ and $\\varphi$ is a homomorphism, $\\varphi(e_G) \\varphi(e_G) = \\varphi(e_G e_G) = \\varphi(e_G)$.\n3. Since $e_H$ is the identity of $H$, $\\varphi(e_G) = \\varphi(e_G) e_H$.\n4. By steps 2 and 3, $\\varphi(e_G) \\varphi(e_G) = \\varphi(e_G) e_H$. By the [[cancellation laws|0VSNX4P2Q0]], $\\varphi(e_G) = e_H$.\n5. Let $a \\in G$. Since $a a^{-1} = e_G$, step 4 gives $\\varphi(a) \\varphi(a^{-1}) = \\varphi(a a^{-1}) = \\varphi(e_G) = e_H$.\n6. By step 5 and the [[uniqueness of inverses|0D9NDGE9KP]], $\\varphi(a^{-1}) = \\varphi(a)^{-1}$.\n7. Steps 4 and 6 are the two claims of the statement.\n\n## Idea\n\nThe identity is the only element $x$ of a group with $xx = x$, and $\\varphi(e_G)$ has this property. Once the identity is preserved, $\\varphi(a^{-1})$ is a right inverse of $\\varphi(a)$, and inverses in a group are unique.\n",
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