The Second Group Isomorphism Theorem

# The Second Group Isomorphism Theorem

Recall from The First Group Isomorphism Theorem page that if $G$ and $H$ are homomorphism groups with homomorphism $\phi : G \to H$ then:

(1)\begin{align} \quad G / \ker (\phi) \cong \phi (G) \end{align}

We are now ready to prove the second group isomorphism theorem.

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

**Proof of a)**Since $H$ and $N$ are subgroups of $G$ we have that their intersection, $H \cap N$ is also a subgroup of $G$. We want to show that $H \cap N$ is a normal subgroup of $H$.

- Now, since $N$ is a normal subgroup of $G$ we have that for all $n \in N$ and for all $g \in G$ that $gng^{-1} \in N$.

- Let $x \in H \cap N$. Since $H$ is a subgroup of $G$ and $H \cap N \subseteq N$ we have that for all $g \in H$ that $gxg^{-1} \in H$. Also, since $x \in H \cap N$ we have that $x \in N$. And since $N$ is a normal subgroup of $G$ we have that for all $g \in H$ that $gxg^{-1} \in N$.

- So for all $x \in H \cap N$ and for all $g \in H$ we have that:

\begin{align} \quad gxg^{-1} \in H \cap N \end{align}

- This shows that $H \cap N$ is a normal subgroup of $H$, that is, $H \cap N \trianglelefteq H$. $\blacksquare$

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

\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:

\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:

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

- And lastly, if $hnN \in (HN)/N$ then $hnN = hN$. So the element $h \in H$ is such that $\phi(h) = hN = hnN \in (HN)/N$. This show that $\phi$ is a surjective function and so:

\begin{align} \quad \phi (H) = (HN)/N \end{align}

- By the first group isomorphism theorem we have that:

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