[proofplan]
We prove equality by double inclusion using the defining minimality property of generated subgroups. First, $\varphi(\langle S\rangle)$ is a subgroup of $H$ containing $\varphi(S)$, so it contains the subgroup generated by $\varphi(S)$. Conversely, the preimage of $\langle \varphi(S)\rangle$ under $\varphi$ is a subgroup of $G$ containing $S$, so it contains $\langle S\rangle$, which gives the reverse inclusion after applying $\varphi$ elementwise.
[/proofplan]
[step:Show that the image of $\langle S\rangle$ contains the subgroup generated by $\varphi(S)$]
Let $A := \langle S\rangle$, and define the set image
\begin{align*}
\varphi(A) := \{\varphi(a) : a \in A\} \subset H.
\end{align*}
We first prove that $\varphi(A)$ is a subgroup of $H$. Since $1_G \in A$ and $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism), $\varphi(1_G) = 1_H$, hence $1_H \in \varphi(A)$. If $u, v \in \varphi(A)$, then there exist $a, b \in A$ such that $u = \varphi(a)$ and $v = \varphi(b)$. Because $A$ is a subgroup of $G$, $ab^{-1} \in A$. Also, since $\varphi$ is a homomorphism,
\begin{align*}
uv^{-1} = \varphi(a)\varphi(b)^{-1} = \varphi(a)\varphi(b^{-1}) = \varphi(ab^{-1}).
\end{align*}
Thus $uv^{-1} \in \varphi(A)$, so $\varphi(A)$ is a subgroup of $H$.
Since $S \subset A$, every element of $\varphi(S)$ lies in $\varphi(A)$. Therefore $\varphi(A)$ is a subgroup of $H$ containing $\varphi(S)$. By the defining minimality of the generated subgroup $\langle \varphi(S)\rangle$, we obtain
\begin{align*}
\langle \varphi(S)\rangle \subset \varphi(A) = \varphi(\langle S\rangle).
\end{align*}
[guided]
Let $A := \langle S\rangle$. The goal of this step is to compare $\langle \varphi(S)\rangle$ with $\varphi(A)$ by using the smallest-subgroup property of generated subgroups. To use that property, we must prove two things: $\varphi(A)$ is a subgroup of $H$, and $\varphi(A)$ contains $\varphi(S)$.
Define
\begin{align*}
\varphi(A) := \{\varphi(a) : a \in A\} \subset H.
\end{align*}
We verify the [subgroup criterion](/theorems/932). Since $A$ is a subgroup of $G$, it contains $1_G$. Since $\varphi$ is a group homomorphism, $\varphi(1_G) = 1_H$, so $1_H \in \varphi(A)$.
Now take arbitrary elements $u, v \in \varphi(A)$. By definition of set image, there exist $a, b \in A$ such that $u = \varphi(a)$ and $v = \varphi(b)$. Since $A$ is a subgroup, $ab^{-1} \in A$. The homomorphism property gives $\varphi(ab^{-1}) = \varphi(a)\varphi(b^{-1})$, and the inverse-preservation property of homomorphisms gives $\varphi(b^{-1}) = \varphi(b)^{-1}$. Therefore
\begin{align*}
uv^{-1} = \varphi(a)\varphi(b)^{-1} = \varphi(a)\varphi(b^{-1}) = \varphi(ab^{-1}).
\end{align*}
Because $ab^{-1} \in A$, the final element $\varphi(ab^{-1})$ lies in $\varphi(A)$. Hence $uv^{-1} \in \varphi(A)$, and $\varphi(A)$ is a subgroup of $H$.
It remains to check that this subgroup contains the generating set $\varphi(S)$. Since $S \subset A$, if $y \in \varphi(S)$, then $y = \varphi(s)$ for some $s \in S$, and that same element $s$ lies in $A$. Thus $y \in \varphi(A)$. Therefore $\varphi(A)$ is a subgroup of $H$ containing $\varphi(S)$.
The subgroup $\langle \varphi(S)\rangle$ is, by definition, the smallest subgroup of $H$ containing $\varphi(S)$. Since $\varphi(A)$ is one such subgroup, minimality gives
\begin{align*}
\langle \varphi(S)\rangle \subset \varphi(A) = \varphi(\langle S\rangle).
\end{align*}
[/guided]
[/step]
[step:Pull back $\langle \varphi(S)\rangle$ to obtain the reverse inclusion]
Define
\begin{align*}
K := \{g \in G : \varphi(g) \in \langle \varphi(S)\rangle\}.
\end{align*}
We prove that $K$ is a subgroup of $G$. Since $1_H \in \langle \varphi(S)\rangle$ and $\varphi(1_G) = 1_H$, we have $1_G \in K$. If $x, y \in K$, then $\varphi(x), \varphi(y) \in \langle \varphi(S)\rangle$. Since $\langle \varphi(S)\rangle$ is a subgroup of $H$,
\begin{align*}
\varphi(x)\varphi(y)^{-1} \in \langle \varphi(S)\rangle.
\end{align*}
Using the homomorphism and inverse-preservation properties of $\varphi$,
\begin{align*}
\varphi(xy^{-1}) = \varphi(x)\varphi(y^{-1}) = \varphi(x)\varphi(y)^{-1}.
\end{align*}
Hence $\varphi(xy^{-1}) \in \langle \varphi(S)\rangle$, so $xy^{-1} \in K$. Thus $K$ is a subgroup of $G$.
For every $s \in S$, we have $\varphi(s) \in \varphi(S) \subset \langle \varphi(S)\rangle$, so $s \in K$. Hence $K$ is a subgroup of $G$ containing $S$. By the defining minimality of $\langle S\rangle$,
\begin{align*}
\langle S\rangle \subset K.
\end{align*}
Therefore, if $a \in \langle S\rangle$, then $a \in K$, so $\varphi(a) \in \langle \varphi(S)\rangle$. This proves
\begin{align*}
\varphi(\langle S\rangle) \subset \langle \varphi(S)\rangle.
\end{align*}
[/step]
[step:Conclude equality from the two inclusions]
The first step proved
\begin{align*}
\langle \varphi(S)\rangle \subset \varphi(\langle S\rangle).
\end{align*}
The second step proved
\begin{align*}
\varphi(\langle S\rangle) \subset \langle \varphi(S)\rangle.
\end{align*}
By double inclusion,
\begin{align*}
\varphi(\langle S\rangle) = \langle \varphi(S)\rangle.
\end{align*}
This is the desired equality.
[/step]
