# If the size of 2 subgroups of G are coprime then why is their intersection is trivial?

Let H and K be subgroups of G, with size p and q respectively, where p and q are coprime, how can we show that H intersect K is {e} where e is the identity element in G

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Hint: What structure does $H\cap K$ have? Use Lagrange's theorem. – Aaron May 31 '11 at 16:59
@Araon thanks for the tip, sorted now, much easier than I thought. – Freeman May 31 '11 at 17:12

$H\cap K$ is a subgroup of $H$. It is also a subgroup of $K$. Now use Lagrange's Theorem.

More generally, if $H$ and $K$ are finite, then $|H\cap K|$ divides $\gcd(|H|,|K|)$. This is a simple corollary of that more general result.

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cheers, that's much clearer – Freeman May 31 '11 at 17:05

Note that $H \cap K$ is a subgroup of $H$ and $K$. Hence, by Lagrange's theorem, the size of $H \cap K$ divides $p$ and $q$. $(p,q) = 1 \implies H \cap K = \{e\}$

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Awesome thanks! – Freeman May 31 '11 at 17:05

LHS, as a bonus: try to solve a similar (dual) problem: Let H and K be subgroups of G, with index p and q respectively, where p and q are coprime, show that G=HK.

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HINT $\ \$ If $\rm\:A,B,C\:$ have cardinality $\rm\:a,b,c\:$ then, using Lagrange's Theorem, we have
$\$

\begin{array}{ccc} \rm A \:\subset\: B\cap C & \Rightarrow &\rm A \:\subset\: B,C \\
& \phantom{\sum} & \Downarrow \\ \rm a\ |\ gcd(b,c) & \Leftarrow &\rm\ a\ \ |\ \ b,\ c \\
\end{array}
$\$

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