First Isomorphism Theorem for Groups If \( \phi : G \rightarrow H \) is a homomorphism then
\[ G / \operatorname { ker } ( \phi ) \cong \operatorname { im } ( \phi ) \]
Proof. Define a mapping \( f:G / \operatorname { ker } ( \phi ) \rightarrow \operatorname { im } ( \phi ) \) by \( f(a\operatorname { ker } ( \phi ))= \phi (a) \)
The map is well-defined since if \( a\operatorname { ker } ( \phi )=b\operatorname { ker } ( \phi ) \), then \( a^{-1}b \in \operatorname { ker } ( \phi ) \) and
Let \( h \) be an arbitrary element of \( \operatorname { im } ( \phi ) \), then \( \exists g \in G : \phi(g)=h \). Since \( f(g\operatorname { ker } ( \phi ))=\phi(g)=h \), \( f:G / \operatorname { ker } ( \phi ) \rightarrow \operatorname { im } ( \phi ) \) is surjective. It is also injective because \( f(a\operatorname { ker } ( \phi ))=f(b\operatorname { ker } ( \phi )) \implies \phi(a)=\phi(b) \implies \phi(a)\phi(b^{-1})=e_H \)\( \implies ab^{-1}\in \operatorname { ker } ( \phi ) \implies a\operatorname { ker } ( \phi )=b\operatorname { ker } ( \phi ) \)
\( f:G / \operatorname { ker } ( \phi ) \rightarrow \operatorname { im } ( \phi ) \) is a homomorphism since
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