[proofplan]
The criterion separates the two parts of being an isomorphism: injectivity and surjectivity. In one direction, an isomorphism is bijective, so its image is all of $H$, and injectivity forces only the identity of $G$ to map to the identity of $H$. In the converse direction, the kernel-equals-identity hypothesis gives injectivity by applying $\varphi$ to $g^{-1}g'$, while the image hypothesis gives surjectivity. Finally, if the definition of isomorphism is taken to require a homomorphic inverse, we verify directly that the inverse of the resulting bijective homomorphism is again a homomorphism.
[/proofplan]
[step:Extract the kernel and image conditions from an isomorphism]
Assume first that $\varphi: G \to H$ is a [group isomorphism](/page/Group%20Isomorphism). Then $\varphi$ is bijective. Since $\varphi$ is surjective, every element of $H$ has the form $\varphi(g)$ for some $g \in G$, and hence
\begin{align*}
\operatorname{im}\varphi = H.
\end{align*}
It remains to compute the kernel. The kernel is
\begin{align*}
\ker\varphi = \{g \in G : \varphi(g) = 1_H\}.
\end{align*}
Because $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism), $\varphi(1_G) = 1_H$, so $1_G \in \ker\varphi$. Conversely, if $g \in \ker\varphi$, then $\varphi(g) = 1_H = \varphi(1_G)$. Since $\varphi$ is injective, $g = 1_G$. Therefore
\begin{align*}
\ker\varphi = \{1_G\}.
\end{align*}
[/step]
[step:Use the trivial kernel condition to prove injectivity]
Assume conversely that
\begin{align*}
\ker\varphi = \{1_G\} \quad \text{and} \quad \operatorname{im}\varphi = H.
\end{align*}
We prove that $\varphi$ is injective. Let $g,g' \in G$ and suppose that $\varphi(g) = \varphi(g')$. Since $\varphi$ is a group homomorphism,
\begin{align*}
\varphi(g^{-1}g') = \varphi(g^{-1})\varphi(g') = \varphi(g)^{-1}\varphi(g').
\end{align*}
Using $\varphi(g) = \varphi(g')$, this becomes
\begin{align*}
\varphi(g^{-1}g') = \varphi(g)^{-1}\varphi(g) = 1_H.
\end{align*}
Thus $g^{-1}g' \in \ker\varphi$. Since $\ker\varphi = \{1_G\}$, we get $g^{-1}g' = 1_G$, and multiplying on the left by $g$ gives $g' = g$. Hence $\varphi$ is injective.
[guided]
We want to prove injectivity, so we start with two arbitrary elements $g,g' \in G$ satisfying $\varphi(g) = \varphi(g')$ and prove $g = g'$. The kernel hypothesis only tells us about elements that map to $1_H$, so the main idea is to turn the equality $\varphi(g) = \varphi(g')$ into a statement that some element of $G$ maps to $1_H$.
The correct element to test is $g^{-1}g' \in G$. Since $\varphi$ is a group homomorphism, it preserves multiplication and inverses. Therefore
\begin{align*}
\varphi(g^{-1}g') = \varphi(g^{-1})\varphi(g') = \varphi(g)^{-1}\varphi(g').
\end{align*}
Now use the assumed equality $\varphi(g) = \varphi(g')$. Substituting $\varphi(g)$ for $\varphi(g')$ gives
\begin{align*}
\varphi(g^{-1}g') = \varphi(g)^{-1}\varphi(g) = 1_H.
\end{align*}
By the definition of the kernel,
\begin{align*}
g^{-1}g' \in \ker\varphi.
\end{align*}
The hypothesis says $\ker\varphi = \{1_G\}$, so this membership forces
\begin{align*}
g^{-1}g' = 1_G.
\end{align*}
Multiplying this equality on the left by $g$ gives $g' = g$. Since every pair $g,g' \in G$ with $\varphi(g) = \varphi(g')$ must be equal, $\varphi$ is injective.
[/guided]
[/step]
[step:Use the image condition to prove surjectivity]
The equality $\operatorname{im}\varphi = H$ is exactly the statement that every element of $H$ lies in the image of $\varphi$. Thus, for every $h \in H$, there exists $g \in G$ such that $\varphi(g) = h$. Therefore $\varphi$ is surjective.
[/step]
[step:Construct the inverse homomorphism and conclude isomorphism]
From the previous two steps, $\varphi: G \to H$ is a bijective group homomorphism. If group isomorphism is defined as a bijective homomorphism, this already proves that $\varphi$ is a group isomorphism.
If group isomorphism is defined as a homomorphism admitting a homomorphic inverse, define the inverse map
\begin{align*}
\psi: H &\to G
\end{align*}
by letting $\psi(h)$ be the unique element $g \in G$ such that $\varphi(g) = h$. This is well-defined because $\varphi$ is bijective. We verify that $\psi$ is a group homomorphism. Let $h_1,h_2 \in H$, and define $g_1,g_2 \in G$ by
\begin{align*}
\psi(h_1) = g_1 \quad \text{and} \quad \psi(h_2) = g_2.
\end{align*}
Then $\varphi(g_1) = h_1$ and $\varphi(g_2) = h_2$. Since $\varphi$ is a homomorphism,
\begin{align*}
\varphi(g_1g_2) = \varphi(g_1)\varphi(g_2) = h_1h_2.
\end{align*}
By the defining uniqueness of $\psi(h_1h_2)$, this implies
\begin{align*}
\psi(h_1h_2) = g_1g_2 = \psi(h_1)\psi(h_2).
\end{align*}
Thus $\psi$ is a group homomorphism. Hence $\varphi$ has a homomorphic inverse and is a group isomorphism. Combining this with the first direction proves the equivalence.
[/step]
