any maths geeks out there?

[ajva]

Erm...

I am right in thinking, am I not, that a group of order 1722 can have only one Sylow 7-subgroup? Or have I missed something?

It's just that the next bit tells me to use a counting argument to show that if it has more than one then it can have only one Sylow 41-subgroup. It seems like the second question is supposed to be theoretical but I just wanted to make sure. :o(

Any comments welcome, although I don't expect I'm going to be snowed under with responses.

Comments

[ciphergoth.livejournal.com]

According to "Sets and Groups", a Sylow subgroup is just any subgroup of order p^a where p^(a+1) doesn't divide |G|. So any 7-subgroup is a Sylow 7-subgroup; it doesn't have to be constructed the way Sylow's theorem prescribes.

Given that, I don't yet see why no group of order 1722 can have more than one subgroup of order 7, but I'm still thinking about it...

[ajva.livejournal.com]

Well, the thing is,

1722 = 2 x 3 x 7 x 41, its prime factorisation.

It you're taking a basic approach (which I am), then the number of Sylow subgroups divides 1722/7 i.e. 246, and is congruent to 1 modulo 7. But there isn't a value which satisfies both of these except 1 (or is there???). So there is only one Sylow 7-subgroup.

Um... is that right?

[ciphergoth.livejournal.com]

But there isn't a value which satisfies both of these except 1 (or is there???)

Yes there is: 246 = 1 (mod 7).

D'oh! (A Homer Simpson moment)

[ajva.livejournal.com]

Bingo!

Cheers Paul. ;o)

Also, it makes the next question a piece of piss.

A. x