Find the number of subgroups of index of the elementary abelian group .
The answer should be but i don't see how to approach this problem.
Note first that is a vector space over . Moreover, if then it's also trivially a subspace of since it's closed under addition. Thus, the subgroups of index , which are precisely the subgroups of order , which are precisely the subspaces of dimension . Now, using the common formula that the number of subspaces of an -dimensional -space is (which is just a simple counting argument about choosing linearly independent subsets of size --it's on page 412 of Dummit and Foote for example) and pluggin in gives the desired result.
thank you drexel but your solution is too advanced for me... actually i wanted to solve the question given on pg 168(question 8(b)) of dummit and foote. i had posted the question on this thread http://www.mathhelpforum.com/math-he...reply&t=186533 . then i took a special case of this problem which is "prove that number of subgroups of order in is equal to the number of subgroups of index in ." Now i was able to prove that has subgroups of order .
I want to solve the question using only the material covered till pg. 168. Your solution uses Vector spaces which is given on pg. 388 and i have not yet read that..
Use the fourth isomorphism theorem to reduce the problem to the case of elementary abelian groups and then take a look at my solution. I mean, not to be pushy, but the only alternative I can think of is a nasty argument using the Orbit Stabilizer theorem. It's easier to just use what I said, and realize that math isn't an insular subject--subjects are supposed to be intermingled.