Show that $H \cap K$ is an abelian subgroup

Let $G$ be a finite, multiplicatively written group and let $H$ be a subgroup of $G$ of order $33$ and let $K$ be a subgroup of $G$ of order $77$. Show that $H ∩ K$ is an abelian subgroup of $G$.

I think I need to use Lagrange's theorem here, I'm not sure how to apply it to this question though. Or maybe $\left\lvert H \cap K \right\rvert$ divides $33$ and $77$ and then go somewhere with that?

The order of $H \cap K$ must divide both $33$ and $77$, hence it must divide their greatest common divisor, which turns out to be $11$. This is because $H \cap K$ is a subgroup of both $H$ and $K$.

So, there are two possibilities:

1. $|H \cap K| = 1$, and you know that this is the trivial group, which is obviously abelian

2. $|H \cap K| = 11$, and you should know that a group of prime order is cyclic.

In both cases, $H \cap K$ is abelian, so we can conclude the proof.

$H \cap K$ is a subgroup, then your order ir 1 or 11 (By Lagrange), in two cases is a prime number, then the group is cyclic and abelian.

• Just to be picky: $1$ isn't prime, so in only one case the order is prime. [But the other case of order 1 is obvious anyway..] May 9 '16 at 19:23

You're on the right track.

Lagrange's theorem tells us that the order of $H \cap K$ divides both $33$ and $77$ and so divides their gcd, which is $11$.

Since $11$ is prime, the order of $H \cap K$ is either $1$ or $11$ and so $H \cap K$ is cyclic, hence abelian.