Third isomorphism theorem

Dependencies:

  1. Factor group
  2. Normal correspondence theorem
  3. gH = H iff g in H
  4. First isomorphism theorem

If $K$ is a normal subgroup of $G$ and $N$ is a normal subgroup of $K$, then $(G/N)/(K/N) \cong G/K$.

Proof

Let $\psi: G/N \mapsto G/K$, where $\psi(gN) = gK$.

\begin{align} & g_1N = g_2N \\ &\Rightarrow (g_2^{-1}g_1)N = N \\ &\Rightarrow g_2^{-1}g_1 \in N \\ &\Rightarrow g_2^{-1}g_1 \in K \tag{$N \subseteq K$} \\ &\Rightarrow g_2^{-1}g_1K = K \\ &\Rightarrow g_1K = g_2K \\ &\Rightarrow \psi(g_1N) = \psi(g_2N) \end{align}

Therefore, $\psi$ is well-defined.

\begin{align} & \psi((g_1N)(g_2N)) \\ &= \psi(g_1g_2N) \\ &= g_1g_2K \\ &= (g_1K)(g_2K) \\ &= \psi(g_1N)\psi(g_2N) \end{align}

Therefore, $\psi$ is a homomorphism.

\begin{align} & \psi(gN) = \operatorname{id}(G/K) = K \\ &\iff gK = K \\ &\iff g \in K \\ &\iff gN \in K/N \end{align}

Therefore, $\operatorname{ker}(\psi) = K/N$.

\begin{align} & \psi(G/N) \\ &= \{ \psi(gN): gN \in G/N \} \\ &= \{ gK: g \in G \} \\ &= G/K \end{align}

Therefore, $\psi(G/N) = G/K$.

Since $\psi: G/N \mapsto G/K$ is a homomorphism, by the first isomorphism theorem, we get

\[ \psi(G/N) \cong (G/N)/\operatorname{ker}(\psi) \Rightarrow G/K \cong (G/N)/(K/N) \]

Dependency for: None

Info:

Transitive dependencies:

  1. Group
  2. Coset
  3. For subset H of a group, a(bH) = (ab)H and (aH)b = a(Hb)
  4. Homomorphism on groups
  5. Mapping of power is power of mapping
  6. Identity of a group is unique
  7. Subgroup
  8. Normal Subgroup
  9. Product of normal cosets is well-defined
  10. Factor group
  11. Inverse of a group element is unique
  12. gH = H iff g in H
  13. Conditions for a subset to be a subgroup
  14. Condition for a subset to be a subgroup
  15. Homomorphic mapping of subgroup of domain is subgroup of codomain
  16. Homomorphic mapping and inverse mapping of normal subgroup is normal
  17. First isomorphism theorem
  18. Correspondence theorem
  19. Normal correspondence theorem