Exercise 3.1.24 ($N \unlhd G \Rightarrow N \cap H \unlhd H$)

Question

Prove that if $N \unlhd G$ and $H$ is any subgroup of $G$ then $N\cap H \unlhd H$.

Richard Ganaye · Accepted Answer

\begin{center}
\begin{tikzpicture}

ode (c1) at (4,0) {$ N \cap H$};

ode (c2) at (3,1) {$N$};

ode (c3) at (5,2) {$H$};

ode (c4) at (4,3) {$NH$};

\draw (c1) edge  (c2);
\draw (c1) edge  (c3);
\draw (c3) edge  (c4);
\draw (c2) edge  (c4);
\end{tikzpicture}
\end{center}

\begin{proof} Let $h$ be any element of $H$, and $a \in N \cap H$.

Since $h \in H$ and $a \in H$,  where $H$ is a subgroup of $G$, then $hah^{-1} \in H$.

Moreover, $a \in N$, and $h \in G$, where $N \unlhd G$, thus $hah^{-1} \in N$.

Therefore $hah^{-1} \in N \cap H$.

Since $hah^{-1} \in N \cap H$ for every $h  \in H$ and every $a \in N \cap H$, we obtain
$$N \cap H \unlhd H.$$

\end{proof}
Note: This is part of the proof of the Second Isomorphism Theorem.

Answers