# Group Order a product of primes

• Nov 17th 2010, 10:17 PM
matt.qmar
Group Order a product of primes
Hello!

If there is a group G of order pqr

(for distict primes p, q and r)

with subgroups H and K of G,
where the order of H is pq
and the order of K is qr

I am trying to show that the order of $\displaystyle H \cap K$ must be q.

-----

Not sure where to run with this one?

I thought maybe that an intersection of two subgroups may be a subgroup of either of those subgroups,
ie, $\displaystyle H \cap K$ is a subgroup of H, so the order of $\displaystyle H \cap K$ must divide the order of H,
and similarily,
and also $\displaystyle H \cap K$ is a subgroup of K,
so the order of $\displaystyle H \cap K$ must divide the order of K

but I am not sure if that is valid or not (is the intersection of two subgroups (H and K) a subgroup itself of either of the two orginal subgroups (H or K)?

Any help appreciated!
Thank you!!
• Nov 17th 2010, 11:49 PM
Swlabr
Your working is right (intersecting two subgroups gives a subgroup - if you do not know this, it is a good exercise to prove it), but only goes so far. You have basically shown that either,

$\displaystyle |H\cap K|=q$ or $\displaystyle |H \cap K|=1$.

It is therefore sufficient to prove that there exists some $\displaystyle 1 \neq g \in H\cap K$.

Can you prove that such an element exists?
• Nov 18th 2010, 12:00 AM
matt.qmar
I will try proving that! Perhaps by one of the subgroup tests! Nice result that I did not previously know.

Back to the question at hand though - I can't seem to think of a way to show some there is a non-identity element in the intersection!

I am trying to recall any theorem regarding the fact that both H and K have a common divisor in their orders, so is it the case that they have elements of the same order ( I have a feeling that this is only for Albelian groups, Cauchy's theorem)- and even then, that still doesn't mean that they have the same element!

Thanks.
• Nov 18th 2010, 03:29 AM
tonio
Quote:

Originally Posted by matt.qmar
I will try proving that! Perhaps by one of the subgroup tests! Nice result that I did not previously know.

Back to the question at hand though - I can't seem to think of a way to show some there is a non-identity element in the intersection!

I am trying to recall any theorem regarding the fact that both H and K have a common divisor in their orders, so is it the case that they have elements of the same order ( I have a feeling that this is only for Albelian groups, Cauchy's theorem)- and even then, that still doesn't mean that they have the same element!

Thanks.

Try using that $\displaystyle \displaystyle{|H\cap K|=\frac{|HK|}{|H||K|}}$ and assuming that the intersection is trivial.

Tonio
• Nov 18th 2010, 03:22 PM
matt.qmar
Hey, that works! thank you!

My only question is, how do you know

$\displaystyle {|H\cap K|=\frac{|HK|}{|H||K|}}$

holds?

is that true for all groups H, K and internal direct product HK? can it be simple proved?

Thanks!!
• Nov 18th 2010, 06:18 PM
tonio
Quote:

Originally Posted by matt.qmar
Hey, that works! thank you!

My only question is, how do you know

$\displaystyle {|H\cap K|=\frac{|HK|}{|H||K|}}$

holds?

is that true for all groups H, K and internal direct product HK? can it be simple proved?

Thanks!!

It is true even for subsets H, K of a group G. Try first proving it by yourself, and then google it. It's a pretty standard result.

Tonio
• Nov 19th 2010, 02:59 AM
topspin1617
Wait, isn't that fraction upside down? Should be

$\displaystyle |HK|=\frac{|H||K|}{|H\cap K|}$.
• Nov 19th 2010, 03:49 AM
tonio
Quote:

Originally Posted by topspin1617
Wait, isn't that fraction upside down? Should be

$\displaystyle |HK|=\frac{|H||K|}{|H\cap K|}$.

Of course...well noted!

Tonio
• Nov 19th 2010, 08:37 AM
matt.qmar
Thank you! It is a fairly obvious result, now that I see it that way.