[proofplan] We first use the homomorphism identity on $1_G\cdot_G 1_G$ to show that $\varphi(1_G)$ is an idempotent element of $H$. Multiplying this equality by the inverse of $\varphi(1_G)$ in $H$ forces $\varphi(1_G)$ to be $1_H$. Then, for an arbitrary $g\in G$, we apply the homomorphism identity to $g\cdot_G g^{-1}=1_G$ and use the identity result to identify $\varphi(g^{-1})$ as the inverse of $\varphi(g)$. [/proofplan] [step:Prove that the identity element maps to the identity element] Since $1_G\cdot_G 1_G=1_G$, the homomorphism property gives \begin{align*} \varphi(1_G) &=\varphi(1_G\cdot_G 1_G) \\ &=\varphi(1_G)\cdot_H\varphi(1_G). \end{align*} Let $u:=\varphi(1_G)\in H$. The displayed equality says $u=u\cdot_H u$. Since $H$ is a group, $u$ has an inverse $u^{-1}\in H$. Left-multiplying both sides by $u^{-1}$ in $H$ gives \begin{align*} u^{-1}\cdot_H u &=u^{-1}\cdot_H (u\cdot_H u). \end{align*} By associativity in $H$ and the inverse law, \begin{align*} 1_H &=(u^{-1}\cdot_H u)\cdot_H u \\ &=1_H\cdot_H u \\ &=u. \end{align*} Thus $u=1_H$, and therefore \begin{align*} \varphi(1_G)=1_H. \end{align*} [guided] The identity element in $G$ satisfies $1_G\cdot_G 1_G=1_G$. Since $\varphi:G\to H$ is a [group homomorphism](/page/Group%20Homomorphism), it respects the group operations $\cdot_G$ and $\cdot_H$. Applying the homomorphism property to the product $1_G\cdot_G 1_G$ gives \begin{align*} \varphi(1_G) &=\varphi(1_G\cdot_G 1_G) \\ &=\varphi(1_G)\cdot_H\varphi(1_G). \end{align*} Set $u:=\varphi(1_G)\in H$. The equality just obtained is $u=u\cdot_H u$. We want to prove $u=1_H$, and the group structure on $H$ lets us cancel $u$ from the left. Because $H$ is a group, $u$ has an inverse $u^{-1}\in H$. Left-multiplying the equality $u=u\cdot_H u$ by $u^{-1}$ gives \begin{align*} u^{-1}\cdot_H u &=u^{-1}\cdot_H (u\cdot_H u). \end{align*} Associativity in $H$ allows us to regroup the right-hand side, and the inverse and identity laws in $H$ give \begin{align*} 1_H &=(u^{-1}\cdot_H u)\cdot_H u \\ &=1_H\cdot_H u \\ &=u. \end{align*} Thus $u=1_H$. Since $u$ was defined to be $\varphi(1_G)$, we conclude \begin{align*} \varphi(1_G)=1_H. \end{align*} [/guided] [/step] [step:Prove that inverses map to inverses] Let $g\in G$ be arbitrary. Since $g^{-1}$ is the inverse of $g$ in $G$, we have \begin{align*} g\cdot_G g^{-1}=1_G. \end{align*} Applying $\varphi$ and using the homomorphism property, \begin{align*} \varphi(g)\cdot_H\varphi(g^{-1}) &=\varphi(g\cdot_G g^{-1}) \\ &=\varphi(1_G) \\ &=1_H, \end{align*} where the last equality follows from the previous step. Hence $\varphi(g^{-1})$ is a right inverse of $\varphi(g)$ in the group $H$. Since $H$ is a group, $\varphi(g)$ has an inverse $\varphi(g)^{-1}\in H$. Left-multiplying the equality $\varphi(g)\cdot_H\varphi(g^{-1})=1_H$ by $\varphi(g)^{-1}$ and using associativity, the inverse law, and the identity law in $H$, we obtain \begin{align*} \varphi(g^{-1}) &=1_H\cdot_H\varphi(g^{-1}) \\ &=(\varphi(g)^{-1}\cdot_H\varphi(g))\cdot_H\varphi(g^{-1}) \\ &=\varphi(g)^{-1}\cdot_H(\varphi(g)\cdot_H\varphi(g^{-1})) \\ &=\varphi(g)^{-1}\cdot_H1_H \\ &=\varphi(g)^{-1}. \end{align*} Because $g\in G$ was arbitrary, this holds for every $g\in G$. [guided] Let $g\in G$ be arbitrary. The inverse of $g$ in $G$ is the element $g^{-1}\in G$, so by the inverse law in $G$, \begin{align*} g\cdot_G g^{-1}=1_G. \end{align*} Apply the homomorphism $\varphi:G\to H$ to this equality. The homomorphism property converts the image of the product in $G$ into the product of the images in $H$: \begin{align*} \varphi(g)\cdot_H\varphi(g^{-1}) &=\varphi(g\cdot_G g^{-1}) \\ &=\varphi(1_G) \\ &=1_H, \end{align*} where the last equality is the identity result proved in the previous step. Thus $\varphi(g^{-1})$ is a right inverse of $\varphi(g)$ in $H$. To identify this right inverse with the usual inverse notation, we cancel $\varphi(g)$ explicitly. Since $H$ is a group, $\varphi(g)$ has an inverse $\varphi(g)^{-1}\in H$. Left-multiplying $\varphi(g)\cdot_H\varphi(g^{-1})=1_H$ by $\varphi(g)^{-1}$ and then using associativity, the inverse law, and the identity law in $H$ gives \begin{align*} \varphi(g^{-1}) &=1_H\cdot_H\varphi(g^{-1}) \\ &=(\varphi(g)^{-1}\cdot_H\varphi(g))\cdot_H\varphi(g^{-1}) \\ &=\varphi(g)^{-1}\cdot_H(\varphi(g)\cdot_H\varphi(g^{-1})) \\ &=\varphi(g)^{-1}\cdot_H1_H \\ &=\varphi(g)^{-1}. \end{align*} Since the element $g\in G$ was arbitrary, the equality holds for every $g\in G$. [/guided] [/step]