Let G be a group of order pqr where p,q and r are distinct primes. If H and K are subgroups of G with |H| = pq and |K| = qr prove that |H intersect K| = q
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H intersect K is a subgroup of both H and K. So by Lagrange m = |H intersect K| divides the |H| = pq and |K|=qr.
Since p, q, and r are primes then the only common factor in pq and qr is q. So m = 1 or q.
If m = 1 then no element of K is in H, so G must have at least |H|K| = prq^2 elements (to see why you can use K as coset representatives for G/H - an they are different for different k's in K).
Since G actually as pqr elements then it must be that m = |H intersect K| = q.
Since p, q, and r are primes then the only common factor in pq and qr is q. So m = 1 or q.
If m = 1 then no element of K is in H, so G must have at least |H|K| = prq^2 elements (to see why you can use K as coset representatives for G/H - an they are different for different k's in K).
Since G actually as pqr elements then it must be that m = |H intersect K| = q.
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The trick is to map the subgroups into groups of p, q & r as follows:
H = Hp x Hq
K = Kq x Kr
H = Hp x Hq
K = Kq x Kr