[proofplan]
In each algebraic setting, the proof is the same closure argument with the relevant operations. We take arbitrary elements of the image, write them as values of $\varphi$, and use the defining preservation laws of the homomorphism to move the operation back to the domain. The group case checks identity, products, and inverses; the ring case checks additive subgroup structure, multiplication, and the unit; the vector-space and module cases check addition and scalar multiplication.
[/proofplan]
[step:Show that the image of a group homomorphism is a subgroup]
Let $G$ and $H$ be groups, and let $\varphi: G \to H$ be a [group homomorphism](/page/Group%20Homomorphism). Define
\begin{align*}
\operatorname{im}\varphi := \{h \in H : \text{there exists } g \in G \text{ such that } \varphi(g)=h\}.
\end{align*}
Let $e_G \in G$ and $e_H \in H$ denote the identity elements. Since $\varphi$ is a group homomorphism, $\varphi(e_G)=e_H$, so $e_H \in \operatorname{im}\varphi$.
Let $u,v \in \operatorname{im}\varphi$. By the definition of image, there exist $a,b \in G$ such that $u=\varphi(a)$ and $v=\varphi(b)$. Since $\varphi$ preserves products,
\begin{align*}
uv = \varphi(a)\varphi(b)=\varphi(ab).
\end{align*}
Because $ab \in G$, this shows $uv \in \operatorname{im}\varphi$.
Also, since $\varphi$ preserves inverses,
\begin{align*}
u^{-1}=\varphi(a)^{-1}=\varphi(a^{-1}).
\end{align*}
Because $a^{-1} \in G$, this shows $u^{-1} \in \operatorname{im}\varphi$. Hence $\operatorname{im}\varphi$ contains the identity of $H$, is closed under the group product of $H$, and is closed under inverses in $H$. Therefore $\operatorname{im}\varphi$ is a subgroup of $H$.
[guided]
Let $G$ and $H$ be groups, and let $\varphi: G \to H$ be a group homomorphism. The subset whose subgroup structure we want to prove is
\begin{align*}
\operatorname{im}\varphi := \{h \in H : \text{there exists } g \in G \text{ such that } \varphi(g)=h\}.
\end{align*}
To prove that this subset is a subgroup of $H$, we verify the subgroup operations directly inside $H$.
First, the identity element of $H$ lies in the image. Let $e_G$ be the identity of $G$ and let $e_H$ be the identity of $H$. Since $\varphi$ is a group homomorphism, it sends the identity of $G$ to the identity of $H$, so
\begin{align*}
\varphi(e_G)=e_H.
\end{align*}
Thus $e_H$ is the value of $\varphi$ at an element of $G$, and therefore $e_H \in \operatorname{im}\varphi$.
Next, take arbitrary elements $u,v \in \operatorname{im}\varphi$. The definition of image means that there are elements $a,b \in G$ such that
\begin{align*}
u=\varphi(a)
\end{align*}
and
\begin{align*}
v=\varphi(b).
\end{align*}
Because $\varphi$ preserves the group operation, we have
\begin{align*}
uv=\varphi(a)\varphi(b)=\varphi(ab).
\end{align*}
The product $ab$ belongs to $G$, so $uv$ is again a value of $\varphi$ on an element of $G$. Hence $uv \in \operatorname{im}\varphi$.
Finally, take $u \in \operatorname{im}\varphi$, and choose $a \in G$ with $u=\varphi(a)$. Since $\varphi$ preserves inverses,
\begin{align*}
u^{-1}=\varphi(a)^{-1}=\varphi(a^{-1}).
\end{align*}
The inverse $a^{-1}$ belongs to $G$, so $u^{-1}$ is again in the image. We have verified that $\operatorname{im}\varphi$ contains $e_H$, is closed under multiplication in $H$, and is closed under inverses in $H$. Therefore $\operatorname{im}\varphi$ is a subgroup of $H$.
[/guided]
[/step]
[step:Show that the image of a unital ring homomorphism is a subring containing the codomain unit]
Let $R$ and $S$ be unital rings, and let $\varphi: R \to S$ be a unital ring homomorphism. Define
\begin{align*}
\operatorname{im}\varphi := \{s \in S : \text{there exists } r \in R \text{ such that } \varphi(r)=s\}.
\end{align*}
Since $\varphi$ is unital,
\begin{align*}
\varphi(1_R)=1_S,
\end{align*}
so $1_S \in \operatorname{im}\varphi$.
Let $u,v \in \operatorname{im}\varphi$. Choose $a,b \in R$ such that $u=\varphi(a)$ and $v=\varphi(b)$. Since $\varphi$ preserves addition,
\begin{align*}
u+v=\varphi(a)+\varphi(b)=\varphi(a+b).
\end{align*}
Because $a+b \in R$, we have $u+v \in \operatorname{im}\varphi$.
Since $\varphi$ preserves additive inverses,
\begin{align*}
-u=-\varphi(a)=\varphi(-a).
\end{align*}
Because $-a \in R$, we have $-u \in \operatorname{im}\varphi$.
Since $\varphi$ preserves multiplication,
\begin{align*}
uv=\varphi(a)\varphi(b)=\varphi(ab).
\end{align*}
Because $ab \in R$, we have $uv \in \operatorname{im}\varphi$. Thus $\operatorname{im}\varphi$ is closed under addition, additive inverses, and multiplication in $S$, and it contains $1_S$. Therefore $\operatorname{im}\varphi$ is a subring of $S$ containing $1_S$.
[/step]
[step:Show that the image of a linear map is a vector subspace]
Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $\varphi: V \to W$ be a [linear map](/page/Linear%20Map). Define
\begin{align*}
\operatorname{im}\varphi := \{w \in W : \text{there exists } v \in V \text{ such that } \varphi(v)=w\}.
\end{align*}
Since $\varphi$ is linear,
\begin{align*}
\varphi(0_V)=0_W,
\end{align*}
so $0_W \in \operatorname{im}\varphi$.
Let $u,v \in \operatorname{im}\varphi$ and let $\lambda \in k$. Choose $a,b \in V$ such that $u=\varphi(a)$ and $v=\varphi(b)$. By additivity of $\varphi$,
\begin{align*}
u+v=\varphi(a)+\varphi(b)=\varphi(a+b).
\end{align*}
Because $a+b \in V$, we have $u+v \in \operatorname{im}\varphi$.
By homogeneity of $\varphi$,
\begin{align*}
\lambda u=\lambda \varphi(a)=\varphi(\lambda a).
\end{align*}
Because $\lambda a \in V$, we have $\lambda u \in \operatorname{im}\varphi$. Hence $\operatorname{im}\varphi$ contains $0_W$ and is closed under addition and scalar multiplication by elements of $k$. Therefore $\operatorname{im}\varphi$ is a vector subspace of $W$.
[/step]
[step:Show that the image of a module homomorphism is a submodule]
Let $A$ be a ring, let $M$ and $N$ be left $A$-modules, and let $\varphi: M \to N$ be an $A$-[module homomorphism](/page/Module%20Homomorphism). Define
\begin{align*}
\operatorname{im}\varphi := \{n \in N : \text{there exists } m \in M \text{ such that } \varphi(m)=n\}.
\end{align*}
Since $\varphi$ is additive,
\begin{align*}
\varphi(0_M)=0_N,
\end{align*}
so $0_N \in \operatorname{im}\varphi$.
Let $u,v \in \operatorname{im}\varphi$ and let $\alpha \in A$. Choose $m_1,m_2 \in M$ such that $u=\varphi(m_1)$ and $v=\varphi(m_2)$. By additivity of $\varphi$,
\begin{align*}
u+v=\varphi(m_1)+\varphi(m_2)=\varphi(m_1+m_2).
\end{align*}
Because $m_1+m_2 \in M$, we have $u+v \in \operatorname{im}\varphi$.
By $A$-linearity of $\varphi$,
\begin{align*}
\alpha u=\alpha \varphi(m_1)=\varphi(\alpha m_1).
\end{align*}
Because $\alpha m_1 \in M$, we have $\alpha u \in \operatorname{im}\varphi$. Hence $\operatorname{im}\varphi$ contains $0_N$ and is closed under addition and left scalar multiplication by elements of $A$. Therefore $\operatorname{im}\varphi$ is a submodule of $N$.
[/step]
