Thread: BG Math [Algebra sucks v2.0]    

bigger and better than before. I'm having a hard time with the last problem of this weeks problem set.

Prove as a corollary of Lagranges Thm (Let H be a subgroup of a finite group G. Then |H| divides |G|) that, if n is an integer not dividing |G|, there does not exist a g in G of order n.

I get the theorems but I just don't see the relation without specifically defining G as a finite group with only one generator, g, to take advantage of the fact that If G = <g> is finite then |G| = ord(g). (and I'm not even sure if I'm going in the right direction)

I was also talking to a friend who said it sorta works for him using the union of equivalence relations defining G, but I didn't really understand his logic.

Thanks for any help you can give me.

It sounds like you're on the right track. If g is an element of G, then <g> is a subgroup of G, and the order of <g> is the order of the element g. So what would happen if we had an element g who's order didn't divide G (what would that tell us about the order of <g>? But what does Lagrange's Theorem say about the order of <g> due to the fact that <g> is a subgroup of G?)?

I hear of the occassional chemist or physicist finding it useful in their applications, but for the most part, group theory seems to be useful only to other mathematicians and theoretical physicists.

Of course, the physics we call theoretical today may become applied in the future. In which case, you'll be glad group theory allows the teleportation device on your 19th G ipod to function properly. Of course, by then I will have taken over the world and will force everybody to study advanced math regardless of whether or not it's useful.

It's weird that an Engineer would have even had to sit in a class where group theory was being taught. But then again, at my school you have to study group theory and real analysis to become an elementary school math teacher, so yeah...

Edit: As a matter of fact, Education majors far outnumber math majors in some of the higher level math classes because math education people have to take as much math as an actual theoretical math major lol. I think they litterally take more math than an applied math major.

Edit: I checked, and Real Analysis is only an elective to education majors. But they do have to take Advance Calc/introductory analysis class and Abstract Algebra (but not the Finite Group Theory course since that's rarely offered).

Originally Posted by willriker

I am just struggling to understand why this guy is posting a second thread on group theory. Are you taking a course on abstract algebra?

Yup. My TA is pretty much useless, and my professor is pretty much inaccessible, so I'm turning to the interwebs for homework help. I'm only a few weeks in, but I'm just not that good at algebra. I usually need help being pointed in the right direction. I have two or three problem sets left this quarter so I hopefully won't need help on a problem more than another two times.

I'm taking real analysis and complex analysis too, but I rarely have big problems with those classes so I never post anything about them.

Originally Posted by wWoozie

If g is an element of G, then <g> is a subgroup of G, and the order of <g> is the order of the element g. So what would happen if we had an element g who's order didn't divide G (what would that tell us about the order of <g>? But what does Lagrange's Theorem say about the order of <g> due to the fact that <g> is a subgroup of G?)

If an element g who's order didn't divide G, I would think that ord(g) > |G|. But of course according to Lagrange's Theorem that's a contradiction since ord(<g>) divides |G|.