Third Isomorphism Theorem for Groups

Third Isomorphism Theorem for Groups Let \( G \) be a group and let \( H \) and \( K \) be normal subgroups of \( G \), with \( H \leq K \). Then

1. \( K / H \unlhd G / H \)
2. \( ( G / H ) / ( K / H ) \cong G / K \)

Proof.

1. \( \forall x \in K/H; y \in G/H, \exists k \in K; g \in G : x=kH; y=gH \) and \( yxy^{-1}=(gkg^{-1})H \). Since \( K \) is a normal subgroup of \( G \), \( gkg^{-1} \in K \). Thus, \( yxy^{-1} \in K/H \) and \( K / H \unlhd G / H \)
2. Consider a mapping \( \phi : G/H \rightarrow G/K \) by \( \phi(gH)=gK \)

   The map is well-defined since if \( aH=bH \), then \( a^{-1}b\in H \subseteq K \) and

   \[ \phi(aH)=aK \circ (a^{-1}b) K = bK=\phi(bH) \]

   The map is homomorphic since \( \forall aH, bH \in G/H \),

   \[ \phi(aH)\phi(bH)=(aK)(bK)=(ab)K=\phi((ab)H) \]

   It is easy to tell \( \phi \) is surjective, so \( \operatorname { im } ( \phi )=G/K \).

   \[ \phi(gH)=gK=K \iff g\in K \iff gH \in K/H \implies \operatorname { ker } ( \phi )=K/H \]

   Based on First Isomorphism Theorem, \( ( G / H ) / ( K / H ) \cong G / K \)

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