# What happens when the order of two finite subgroups are relatively prime?

I'm returning from an exam on group-theory and there were 2 questions I couldn't solve (and still can't), so I'm asking here for any hint you could possibly give.

Let G be a group and H and K subgroups such that $|H| = n$, $|K| = m$ and $gcd(n, m) = 1$. Show that $H \cap K = \{e\}$.

I wish I could show you some of my attempts before hand, but they're all rubbish that didn't get me anywhere. Essentially, the only (and last) thing I remembered and thought it could be useful was to see if if H and K are partitions of G. I think I've read something similar somewhere but can't recall where, so, am uncertain about it.

The other question I couldn't solve is, I think, related to this, so I shall try it once I understand this one.

Thanks for taking the time to read! Any tip is appreciated.

• What do you know about the relationship between the order of a group and the order of one of is subgroups? – Qiaochu Yuan Feb 3 '11 at 17:06
• The order of a group is a multiple of the order of its subgroups, by the Lagrange's theorem, right? So, if $|G| = 20$, $|H|$ and $|K|$ could be 4 and 5. But I fail to see a connection, yet. – Marla Feb 3 '11 at 17:11

Let $g \in H\cap K$. By Lagrange's Theorem, the order of g divides m and n, but the greatest common divisor of m and n is 1, so g=e.
• Great answer, and is $g=e$ because $|H\bigcap K|=1? – grayQuant Mar 2 '15 at 2:33 • @grayQuant The order of$g$is 1, i.e.$g^1=g=e\$. – Ben Derrett Mar 2 '15 at 8:52