Find the number of subgroups of index $\displaystyle p$ of the elementary abelian group $\displaystyle E_{p^n}$.
The answer should be $\displaystyle \frac{p^n-1}{p-1}$ but i don't see how to approach this problem.
Note first that $\displaystyle E_{p^n}$ is a vector space over $\displaystyle \mathbb{F}_p$. Moreover, if $\displaystyle H\leqslant E_{p^n}$ then it's also trivially a subspace of $\displaystyle E_{p^n}$ since it's closed under addition. Thus, the subgroups of index $\displaystyle p$, which are precisely the subgroups of order $\displaystyle p^{n-1}$, which are precisely the subspaces of dimension $\displaystyle n-1$. Now, using the common formula that the number of subspaces of an $\displaystyle n$-dimensional $\displaystyle \mathbb{F}_q$-space is $\displaystyle \displaystyle \frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{(q^k-1)(q^{k}-q)\cdots(q^k-q^{k-1})}$ (which is just a simple counting argument about choosing linearly independent subsets of size $\displaystyle k$--it's on page 412 of Dummit and Foote for example) and pluggin in $\displaystyle k=n-1$ 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 $\displaystyle p$ in $\displaystyle E_{p^n}$ is equal to the number of subgroups of index $\displaystyle p$ in $\displaystyle E_{p^n}$." Now i was able to prove that $\displaystyle E_{p^n}$ has $\displaystyle \frac{p^n-1}{p-1}$ subgroups of order $\displaystyle p$.
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.