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## Count the number of cyclic subgroups in an elementary abelian p-group

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.11

Solution: By definition, every nonidentity element of $E_{p^n}$ has order $p$. Thus every nonidentity element generates an order $p$ subgroup. On the other hand, every order $p$ subgroup has $p-1$ generators. Thus the number of distinct order $p$ subgroups is $(p^n-1)/(p-1)$, which is evidently equal to $$\sum_{k=0}^{n-1} p^k.$$