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I'm reading in Dummit's Abstract Algebra the example (3) p. 156, where they prove the claim "for $p$ a prime, the elementary abelian group of order $p^2$, $E$ (i.e. $Z_{p} \times Z_{p}$, where $Z_{p}$ is the cyclic group of order $p$) has exactly $p+1$ subgroups or order $p$".

They start proving this by saying that, since obviously each nonidentity element of $E$ has order $p$ (since it happens for every element of $Z_{p}$), then they generate a cyclic subgroup of order p. Noting that each of this subgroups has exactly $p-1$ generators (since $p$ is prime) and that by Lagrange's Theorem distinct subgroups of order $p$ intersect trivially, they deduce there are $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$.

I agree with the fact we have found $p+1$ subgroups of order $p$, but I don't know why they suppose they are all the subgroups of order $p$.

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    $\begingroup$ Each of those $p+1$ subgroups has $p-1$ elements of order $p$; those $p^2-1$ elements and the 1 identity element comprise all the elements in a group of order $p^2$ $\endgroup$ – J. W. Tanner Jan 13 at 17:15
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An alternative is to think of this in terms of linear algebra. The elementary abelian group of order $p^n$ is an $n$-dimensional vector space over $\mathbb{F}_p$; the subgroups of order $p$ are exactly the $1$-dimensional subspaces. Each one dimensional subspace contains $p$ points, and each nonzero point can function as a basis for the subspace. Moreover, the intersection of two distinct $1$-dimensional subspaces must be trivial (dimension $0$) by dimensional considerations.

Thus, each nonzero element generates a one-dimensional subspace, and every one-dimensional subspace is thus obtained, and we are counting each of them $p-1$ times. Thus, the number of distinct $1$-dimensional subspaces of the elementary abelian group of order $p^n$ is $\frac{p^n-1}{p-1}$.

One can easily generalize this for $k$-dimensional subspaces, i.e. subgroups of order $p^k$: you need to select a linearly independent subset with $k$ elements. The first one can be picked arbitrarily, $p^n-1$ choices; the next one must be outside the one-dimensional subspace that first generates, $p^n-p$ choices; the next $p^n-p^2$ choices, and so on until the $k$th one has $p^n-p^{k-1}$ choices. But then we have to account for the overcount. How many distinct ordered bases does a $k$-dimensional space over $\mathbb{F}_p$ has? Well, using the same counting technique we see that we can pick an ordered basis in $$(p^k-1)(p^k-p)\cdots(p^k-p^{k-1})\quad\text{ways.}$$ Thus, the number of $k$-dimensional subspaces is $$\frac{(p^n-1)(p^n-p)\cdots(p^n-p^{k-1})}{(p^k-1)(p^k-p)\cdots(p^k-p^{k-1})}.$$ In particular, for $n=2$ and $k=1$ we get $$\frac{p^2-1}{p-1} = p+1.$$

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Each of the $(p+1)$ subgroups has order $p$, and they intersect trivially. Thus we count $(p+1)(p-1)+1=p^2$ elements in the group.

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There are $p^2-1$ elements of order $p$.

Each of the non-intersecting subgroups contains $p-1$ of these elements.

Therefore there are precisely $\frac{p^2-1}{p-1}=p+1$ such subgroups.

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  • $\begingroup$ Thank you for your answer, I see two more users have sugested the same approach, maybe I'm being dumb but I still don't get it; I agree there are $p+1$ cyclic subgroups, but I don't know if we can form another group by selecting a few elements in a few of these groups that, as we have seen, form a partition of the group G (and this new group will not be among the $p+1$ groups we have counted). $\endgroup$ – Amelia Jan 17 at 17:05
  • $\begingroup$ Every subgroup has either $1, p$ or $p^2$ elements by Lagrange and all groups of order $p$ are cyclic so there cannot be any other subgroups. $\endgroup$ – S. Dolan Jan 17 at 17:10
  • $\begingroup$ I got it now, thank you so much ! $\endgroup$ – Amelia Jan 17 at 17:24
  • $\begingroup$ Glad to be of help. $\endgroup$ – S. Dolan Jan 17 at 17:25

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