Exercise 3.2.8 (If $\mathrm{g.c.d.}(|H|, |K|) = 1$ then $H \cap K = 1$)

Question

Prove that if $H$ and $K$ are finite subgroups of $G$ whose orders are relatively prime then $H \cap K = 1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
For every $x \in H \cap K$, by Lagrange's Theorem and Corollary 10, $d = |x|$ is a divisor of $m = |H|$ and $n =|K|$. Therefore $d \mid m \wedge n = 1$, therefore $d = 1$, so $x = 1$. This shows that
$$H \cap K = \{1\}.$$

\end{proof}

Exercise 3.2.8 (If $\mathrm{g.c.d.}(|H|, |K|) = 1$ then $H \cap K = 1$)

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Exercise 3.2.8 (If $\mathrm{g.c.d.}(|H|, |K|) = 1$ then $H \cap K = 1$)

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