Thread: 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 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, is a subgroup of H, so the order of must divide the order of H,
and similarily,
and also is a subgroup of K,
so the order of 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!!

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,

or .

It is therefore sufficient to prove that there exists some .

Can you prove that such an element exists?

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.

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 and assuming that the intersection is trivial.

Tonio

Hey, that works! thank you!

My only question is, how do you know

holds?

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

Thanks!!

Originally Posted by matt.qmar

Hey, that works! thank you!

My only question is, how do you know

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

Wait, isn't that fraction upside down? Should be

.

Originally Posted by topspin1617

Wait, isn't that fraction upside down? Should be

.

Of course...well noted!

Tonio

Thank you! It is a fairly obvious result, now that I see it that way.