# Let G be a group and let H,K be subgroups of G where |H|=12 and |K|=5. Then the intersection of H and K = {e}.

The question is as stated in the title. I used LaGrange's theorem to show that |H|||G| and |K|||G| so 12||G| and 5||G|, and that 12 and 5 are relatively prime. I'm not sure if this has gotten me closer to solving the problem, as I am stuck. Any help is greatly appreciated.

NOTE: This problem provides does not specifically define e, so I'm attempting to prove this assuming e is the identity of G.

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Hint:

Consider $|H\cap K|\,|\,|H|$ and $|H\cap K|\,|\,|K|$.

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Why am I allowed to say that? –  Heath Huffman Mar 11 '14 at 17:53
By LaGrange's theorem, we have $|H\cap K|\,|\,|H|$. –  Babak Miraftab Mar 11 '14 at 17:54
Okay, so since |H∩K|||H| and |H∩K||K|, and |H| and |K| are relatively prime, then |H∩K| = 1 = e the identity of G. But how do we know 1 = e? –  Heath Huffman Mar 11 '14 at 17:58
$o(x)=1$ if and only if $x=e$. Some books use e instead of 1. Note that $e$ is the identity element. –  Babak Miraftab Mar 11 '14 at 18:00

Let $x\in H\cap K$ then by Lagrange theorem the order of $x$ divides the two coprime orders: $|H|$ and $|K|$ so $o(x)=1$ and then $x=e$. Conclude.

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We know that $H\cap K$ has at least the element $e$. suppose that $H\cap K$ has another element $g$ so that $g\in H$ and $g\in K$. the $|H\cap K|$ divides the order of $H$ and $K$ but $5$ and $12$ don't have common divisors. So the only element in $H\cap K$ is $e$.

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