# Calculating the number of subgroups of $\mathbb{Z}/p\mathbb{Z}\oplus\mathbb{Z}/p\mathbb{Z}$

Let $p$ be a prime. I want to find the number of subgroups of the group $$G=\mathbb{Z}/p\mathbb{Z}\oplus\mathbb{Z}/p\mathbb{Z}$$.

My idea:

(1) The main problem is to find subgroups of order $p$. Any subgroup of order $p$ is of the form $$H=\langle(a,b)\rangle$$ where one of $a$ or $b$ is nonzero.

(2) For $0\leq i\leq p-1$, define the subgroup $$H_i=\langle(1,i)\rangle.$$ Define the subgroup $$K=\langle (0,1)\rangle.$$ Then the subgroups $K, H_0,\dots, H_{p-1}$ are distinct subgroups of order $p$.

(3) We claim that $$G=\left(\bigcup_{i=0}^{p-1}H_i\right)\bigcup K.$$ This follows from calculating the cardinality of the union. The subgroups $K, H_0,\dots, H_{p-1}$ are distinct subgroups of order $p$. Any two distinct subgroups of order $p$ intersect only at $(0,0)$. There are $p+1$ subgroups in the union and each of them contains $p-1$ nontrivial elements. So the cardinality of the union is $$\left|\left(\bigcup_{i=0}^{p-1}H_i\right)\bigcup K\right|=(p+1)(p-1)+1=p^2=|G|$$

(4) Since $$G=\left(\bigcup_{i=0}^{p-1}H_i\right)\bigcup K$$ if $\langle(a,b)\rangle$ is any subgroup of order $p$, then $\langle(a,b)\rangle$ is already considered in the union. Therefore $K, H_0,\dots, H_{p-1}$ are precisely all the distinct subgroups of order $p$.

(5) So the total number of subgroups of order $p$ is $p+1$. Hence the total number of subgroups is $p+3$.

Is my calculation correct? What are the other methods to solve this problem?

The argument in (3) and (4) is incomplete. You go from "the union of all these subgroups is all of $G$" to concluding "we must have found all the subgroups of order $p$," but don't explain why.

Indeed cardinality does not seem very relevant to me. Your idea of splitting subgroups of order $p$ into those of two types is a good one - it is an elementary version of the Schubert cell decomposition for projective spaces, $\mathbb{P}^n=\mathbb{A}^n\sqcup\mathbb{A}^{n-1}\sqcup\cdots\sqcup\mathbb{A}^1\sqcup\mathbb{A}^0$. (Don't worry about understanding that, I'm just saying your idea is one used in advanced math.)

To see that you've found every subgroup of order $p$, one can do so directly. Every such subgroup is cyclic, so generated by some $(a,b)$, and $\langle (a,b)\rangle$ is already in your list of found subgroups (with two cases depending on if $a=0$ or $a\ne0$).

Here is another argument that works more generally but you may find too abstract.

The subgroups of $\mathbb{Z}_p^2$ of size $p$ are cyclic, and this is a special case of counting $1$-dim subspaces of a vector space $V$ over a finite field $k$. Let $\mathbb{P}(V)$ be the collection of all $1$-dim subspaces of $V$ and write $V^{\times}$ for the subset of nonzero vectors in $V$. Every nonzero vector $v$ generated a $1$-dim subspace $\langle v\rangle=kv$, so there is a function $V^{\ast}\to \mathbb{P}(V)$ given by $v\mapsto\langle v\rangle$. Any fiber of this map is the set of nonzero vectors of a $1$-dim subspaces, of which there are $|k^{\times}|$-many. Therefore, the number of $1$-dim subspaces is $|V^{\times}|/|k^{\times}|$.

Explicitly, if $k=\mathbb{F}_q$ and $V=\mathbb{F}_q^n$ then this is $(q^n-1)/(q-1)=q^{n-1}+\cdots+q+1$ (which matches the decomposition $\mathbb{P}^{n-1}=\mathbb{A}^{n-1}\sqcup\cdots\sqcup\mathbb{A}^1\sqcup\mathbb{A}^0$).

This should not be surprising if you are familiar with group actions, since the $1$-dim subspaces minus the origin are precisely the orbits of $k^{\times}$ acting on $V^{\times}$ freely.

• Here is why we have found all the subgroups of order $p$: Since $G$ is the union of all the subgroups listed above, any $(a,b)\in G$ is contained in one of the subgroups in the union (say $H_i$). But since the $(a,b)\in G$ has order $p$, and $|H_i|=p$, we must have $H_i=\langle (a,b)\rangle$ – learning_math Aug 12 '17 at 2:41
• Good. That works. – anon Aug 12 '17 at 2:45
• But I like your idea of looking at $G$ as a vector space over $\mathbb{F}_p$. Then the problem reduces down to finding all the one-dimensional subspace. Thank you. – learning_math Aug 12 '17 at 2:46