Proof of 1

The operation that is preserved by ${\displaystyle \phi }$ is the group operation. If ${\displaystyle a,b\in {\text{im}}(\phi )}$, then there exist elements ${\displaystyle a',b'\in G}$ such that ${\displaystyle \phi (a')=a}$ and ${\displaystyle \phi (b')=b}$. For these ${\displaystyle a}$ and ${\displaystyle b}$, we have ${\displaystyle ab=\phi (a')\phi (b')=\phi (a'b')\in {\text{im}}(\phi )}$ (since ${\displaystyle \phi }$ preserves the group operation), and thus, the closure property is satisfied in ${\displaystyle {\text{im}}(\phi )}$. The identity element ${\displaystyle e\in H}$ is also in ${\displaystyle {\text{im}}(\phi )}$ because ${\displaystyle \phi }$ maps the identity element of ${\displaystyle G}$ to it. Since every element ${\displaystyle a'}$ in ${\displaystyle G}$ has an inverse ${\displaystyle (a')^{-1}}$such that ${\displaystyle \phi ((a')^{-1})=(\phi (a'))^{-1}}$ (because ${\displaystyle \phi }$ preserves the inverse property as well), we have an inverse for each element ${\displaystyle \phi (a')=a}$ in ${\displaystyle {\text{im}}(\phi )}$, therefore, ${\displaystyle {\text{im}}(\phi )}$ is a subgroup of ${\displaystyle H}$.

Proof of 2

Construct a map ${\displaystyle \psi :G/\ker(\phi )\to {\text{im}}(\phi )}$ by ${\displaystyle \psi (a\ker(\phi ))=\phi (a)}$. This map is well-defined, as if ${\displaystyle a\ker(\phi )=b\ker(\phi )}$, then ${\displaystyle b^{-1}a\in \ker(\phi )}$ and so ${\displaystyle \phi (b^{-1}a)=e\Rightarrow \phi (b^{-1})\phi (a)=e}$ which gives ${\displaystyle \phi (a)=\phi (b)}$. This map is an isomorphism. ${\displaystyle \psi }$ is surjective onto ${\displaystyle {\text{im}}(\phi )}$ by definition. To show injectiveness, if ${\displaystyle \psi (a\ker(\phi ))=\psi (b\ker(\phi ))}$, then ${\displaystyle \phi (a)=\phi (b)}$, which implies ${\displaystyle b^{-1}a\in \ker(\phi )}$ so ${\displaystyle a\ker(\phi )=b\ker(\phi )}$.

Finally,

$${\displaystyle \psi ((a\ker(\phi ))(b\ker(\phi )))=\psi (ab\ker(\phi ))=\phi (ab)=\phi (a)\phi (b)=\psi (a\ker(\phi ))\psi (b\ker(\phi ))}$$

hence ${\displaystyle \psi }$ preserves the group operation. Hence ${\displaystyle \psi }$ is an isomorphism between ${\displaystyle G/\ker(\phi )}$ and ${\displaystyle {\text{im}}(\phi )}$, which completes the proof.

The integers modulo n

For each ${\displaystyle n\in \mathbb {N} }$, consider the groups ${\displaystyle \mathbb {Z} }$ and ${\displaystyle \mathbb {Z} _{n}}$ and a group homomorphism ${\displaystyle f:\mathbb {Z} \rightarrow \mathbb {Z} _{n}}$ defined by ${\displaystyle m\mapsto m{\text{ mod }}n}$ (see modular arithmetic). Next, consider the kernel of ${\displaystyle f}$, ${\displaystyle {\text{ker}}(f)=n\mathbb {Z} }$, which is a normal subgroup in ${\displaystyle \mathbb {Z} }$. There exists a natural surjective homomorphism ${\displaystyle \varphi$ :\mathbb {Z} \rightarrow \mathbb {Z} /n\mathbb {Z} } defined by ${\displaystyle m\mapsto m+n\mathbb {Z} }$. The theorem asserts that there exists an isomorphism ${\displaystyle h}$ between ${\displaystyle \mathbb {Z} _{n}}$ and ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$, or in other words ${\displaystyle \mathbb {Z} _{n}\cong \mathbb {Z} /n\mathbb {Z} }$. The commutative diagram is illustrated below.

N/C Theorem

Let ${\displaystyle G}$ be a group with subgroup ${\displaystyle H}$. Let ${\displaystyle C_{G}(H)}$, ${\displaystyle N_{G}(H)}$ and ${\displaystyle {\text{Aut}}(H)}$ be the centralizer, the normalizer and the automorphism group of ${\displaystyle H}$ in ${\displaystyle G}$, respectively. Then, N/C theorem states that ${\displaystyle N_{G}(H)/C_{G}(H)}$ is isomorphic to a subgroup of ${\displaystyle {\text{Aut}}(H)}$.