# Clarification on Proof that “for $p$ a prime, the elementary abelian group of order $p^2$ has exactly $p+1$ subgroups or order $p$”

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 $$p$$.

• 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 $p^2$ – J. W. Tanner Jan 13 at 17:15

## 3 Answers

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 $$\frac{p^2-1}{p-1} = p+1.$$

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.

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.

• 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). – Amelia Jan 17 at 17:05
• 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. – S. Dolan Jan 17 at 17:10
• I got it now, thank you so much ! – Amelia Jan 17 at 17:24
• Glad to be of help. – S. Dolan Jan 17 at 17:25