Criterion for (G/K)/(H/K) to be Isomorphic to (G/H) When H, K ◅ G

Criterion for (G/K)/(H/K) to be Isomorphic to (G/H) When H, K are Normal Subgroups of G

Let $G$ be a group and let $H$ and $K$ be normal subgroups of $G$ with $K \subseteq H$. Since $H$ is a normal subgroup of $G$ and $K$ is a normal subgroup of $G$, the quotient $G/H$ and $G/K$ can be formed.

Also, note that since $K$ is a normal subgroup of $G$ we have that for all $g \in G$, $gKg^{-1} = K$. In particular, since $H \subseteq G$, we have that for all $h \in G$, $hKh^{-1} = K$. So $K$ is a normal subgroup of $H$. Thus the quotient $H/K$ can be formed.

We will now present a theorem which relates the quotient groups $G/H$, $G/K$, and $H/K$ together.

Theorem 1: Let $G$ be a group and let $H$ and $K$ be normal subgroups of $G$ such that $K \subseteq H$. Then $(G/K)/(H/K)$ is isomorphic to $G/H$.

• Proof: Let $\phi: G/K \to G/H$ be defined for all $gK \in G/K$ by:

(1)

\begin{align} \quad \phi(gK) = gH \end{align}

• We first show that $\phi$ is a well-defined map. Let $g, g' \in G$ and suppose that $gK = g'K$. Then $g \in g'K$. Since $K \subseteq H$ we see that $g \in g'H$. Then $gH = g'H$. So regardless of the representative $g$ we choose for the coset $gK$, we always have the same output for $\phi(gK)$, i.e., $\phi$ is well-defined.

• Furthermore, $\phi$ is a homomorphism from $G/K$ to $G/H$ since for all $g_1K, g_2K \in G/K$ we have that:

(2)

\begin{align} \quad \phi([g_1K][g_2K]) = \phi(g_1g_2K) = g_1g_2H = [g_1H][g_2H] = \phi(g_1K) \phi(g_2K) \end{align}

• Observe that:

(3)

\begin{align} \quad \ker (\phi) &= \{ gK \in G/K : \phi(gK) = H \} \\ &= \{ gK \in G/K : gH = H \} \\ &= \{ gK \in G/K : g \in H \} \\ &= H/K \end{align}

• So by The First Group Isomorphism Theorem we have that:

(4)

\begin{align} \quad (G/H)/(H/K) = (G/H)/\ker(\phi) \cong \phi(G/K) \end{align}

• But observe that $\phi$ is surjective, since for all $gH \in G/H$ we have that $gK \in G/K$ is such that $\phi(gK) = gH$, and so $\phi(G/K) = G/H$. Thus from above we have that:

(5)

\begin{align} \quad (G/H)/(H/K) \cong G/H \quad \blacksquare \end{align}