Exercise 3.1.42 (If $H \unlhd G$ , $K \unlhd G$ and $H \cap K = 1$, then $xy = yx$ ($x \in H,\ y\in K$))

Question

Assume both $H$ and $K$ are normal subgroups of $G$ with $H \cap K = 1$. Prove that $xy = yx$ for all $x \in H$ and $y\in K$. [Show $x^{-1}y^{-1} xy \in H \cap K$.]

Richard Ganaye · Accepted Answer

\begin{proof} Since $H \unlhd G$, then
$$x^{-1}y^{-1} xy = x(y^{-1} x y) \in H,$$
because $x \in H$ and $y^{-1} x y = y^{-1} x (y^{-1})^{-1} \in H$.
Similarly, since $K \unlhd G$,
$$x^{-1}y^{-1} xy= (x^{-1}y^{-1} x)y \in K.$$
Therefore $x^{-1}y^{-1} xy \in H \cap K = \{1\}$, thus $x^{-1}y^{-1} xy = 1$, so $xy = yx$.

 If $H$ and $K$ are normal subgroups of $G$ with $H \cap K = 1$, then $xy = yx$ for all $x \in H$ and $y\in K$. 


\end{proof}

Exercise 3.1.42 (If $H \unlhd G$ , $K \unlhd G$ and $H \cap K = 1$, then $xy = yx$ ($x \in H,\ y\in K$))

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Exercise 3.1.42 (If $H \unlhd G$ , $K \unlhd G$ and $H \cap K = 1$, then $xy = yx$ ($x \in H,\ y\in K$))

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