Proving |H intersect K| = q for subgroup H and K in G of order pqr.

[mykayla10]

Homework Statement

Suppose |G| = pqr where p, q, and r are distinct primes. If H is a subgroup of G and K is a subgroup of G with |H| = pq and |K| = qr, then |H intersect K| = q.

Homework Equations

NA

The Attempt at a Solution

I have so far:
Let a be an element of H intersect K. Therefore, |a| divides |H| and divides |K|. Therefore, |a| divides pq and qr. Since p, q, and r are distinct primes, we know that the gcd( pq,qr)=q. Since |a| divides both pq and qr, |a| divides the gcd of these two, so |a| divides q. Since q is prime, |a|= 1 or q.

I am now stuck as to how to prove that |a| is not 1. If I do this, by process of elimination I have proven that |H intersects K|=q.

 

[mathwonk]

let one subgroup act on the other by translation. i.e. show all the cosets of form hK with h in H are different. then think about that. i hope this helps. in general everything in group studying theory is done by these sorts of "actions".

 

[mykayla10]

is there any way I can eliminate |H intersect K|=1 by assuming it does and finding a contradiction? I am not sure I am at the ability level to do what you suggested.