The Second Group Isomorphism Theorem

Recall from The Intersection of a Normal Subgroup with a Subgroup is a Normal Subgroup page that if $G$ is a group and $H$, $N$ are subgroups of $G$ such that $N \triangleleft G$ then $H \cap N \trianglelefteq H$.

With this, we are now reading to prove The Second Group Isomorphism Theorem.

Theorem 1 (The Second Group Isomorphism Theorem): Let $G$ be a group and let $H$ and $N$ be subgroups of $G$ such that $N \trianglelefteq G$. Then $H/(H \cap N) \cong (HN)/N$.

From the Theorem referenced at the top of this page we have that since $H$, $N$ are subgroups of $G$ and $N \trianglelefteq H$ that $H \cap N \trianglelefteq H$, and so the quotient $H / (H \cap N)$ is well-defined. Furthermore, recall from The Product HK of Two Subgroups H and K of a Group G page that $HN$ is a group if and only if $HN = NH$. Since $N$ is a normal subgroup of $G$, we have that $gN = Ng$ for all $g \in G$. Since $H \subseteq G$, we have that in particular, $hN = Nh$ for all $h \in H$. So $HN = NH$. Thus, $HN$ is indeed a well-defined group. Furthermore, $N = \{ en : n \in N \} \subseteq HN$ since $e \in H$. Lastly, observe that for all $n \in N$ and for all $x = hn' \in HN$ we have that $xnx^{-1} = hn'nn'^{-1}h^{-1} \in HNH$. But $HN = NH$, so $xnx^{-1} \in HHN = HN$. Since this holds true for all $n \in N$ and for all $x \in HN$, we see that $N \trianglelefteq HN$. So the quotient $(HN)/N$ is well-defined.

• Proof: Define a function $\phi : H \to (HN) / N$ for all $h \in H$ by:

(1)

\begin{align} \quad \phi (h) = hN \end{align}

• We want to show that $\phi$ is a group homomorphism. Let $h_1, h_2 \in H$. Then:

(2)

\begin{align} \quad \phi (h_1h_2) = (h_1h_2)N = (h_1N)(h_2N) = \phi(h_1)\phi(h_2) \end{align}

• So indeed, $\phi$ is a group homomorphism. We now determine the kernel and range of $\phi$. We have that:

(3)

\begin{align} \quad \ker (\phi) &= \{ h \in H : \phi(h) = N \} \\ &= \{ h \in H : hN = N \} \\ &= \{ h \in H : h \in N \} \\ &= H \cap N \end{align}

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

(4)

\begin{align} \quad H/H\cap N = H/\ker(\phi) \cong \phi(H) \end{align}

• But observe that $\phi$ is surjective since if if $hnN \in (HN)/N$ then $hnN = hN$. So the element $h \in H$ is such that $\phi(h) = hN = hnN \in (HN)/N$. Thus $\phi(H) = (HN)/N$, and from above we have that:

(5)

\begin{align} \quad H / (H \cap N) \cong (HN)/N \quad \blacksquare \end{align}