Number of subgroups $H$ of $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p^3}$ with $|H|=p^2$ with prime $p$.
The followings are my trial.
Basically, I am trying to solve this problem via the following lemma
Let $n$ be the number of the elements of order $m$, then there are $\frac{n}{\phi(m)}$ cyclic subgroups of order $m$.
So my problem is nothing but count the number of elements of order $p^2$ in $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p^3}$. i.e., Find number of $x$, $p^2 x =0$ where $x \neq 0 \in \mathbb{Z}_{p^2} \times \mathbb{Z}_{p^3}$
In $\mathbb{Z}_{p^2}$ there are $p^2$ elements such that $p^2 x'=0$ and among them $p$ elements also satisfies $px=0$. Similar thing happens in $\mathbb{Z}_{p^3}$. In this case there are $p$ elements such that $p^2 x'' =0$ and $p^2$ elements such that $px''=0$
So combine I have
$ (p^2-p)\times (p+p^2) + p \times p^2 -1 $
order $p^2$ elements with (order $p$ + order $p^2$) + order $p$ elements with order $p^2$ -1 (0,0)
Am I correct? How one can correctly count the number of $p^2$ in $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p^3}$?
Best Answer
Since , $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p^3}$ is abelian, so, all subgroups must be abelian .
So, first we count the number of cyclic subgroups of order $p^2$ , then abelian subgroups of order $p^2$ which are not cyclic .
If $(a,b)$ be element of $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p^3}$ of order $p^2$
That mean , $|(a,b)|=lcm\{|a|,|b|\}=p^2$ , where , $|a|,|b|$ are order of $a,b$ in $\mathbb{Z}_{p^2}$ ,$\mathbb{Z}_{p^3}$ respectively.
Possible formations are , $lcm\{1,p^{2}\},lcm\{p,p^{2}\},lcm\{p^{2},p^{2}\},lcm\{p^{2},p\},lcm\{p^{2},1\}$
In the case, $lcm\{1,p^{2}\}$ , the number of element of order $1$ in $\mathbb{Z}_{p^2}$ is $1$ And , the number of element of order $p^2$ in $\mathbb{Z}_{p^3}$ is $\phi (p^{2})=p(p-1)$
Number of elements $(a,b)$ in the case of $lcm\{1,p^{2}\}$ is $1\times p(p-1)=p(p-1)$
Similarly, Number of elements $(a,b)$ in the case of $lcm\{p,p^{2}\}$ is $(p-1)\times p(p-1)=p(p-1)^{2}$
Number of elements $(a,b)$ in the case of $lcm\{p^{2},p^{2}\}$ is $p(p-1)\times p(p-1)=p^{2}(p-1)^{2}$
Number of elements $(a,b)$ in the case of $lcm\{p^{2},p\}$ is $ p(p-1) \times (p-1)=p(p-1)^{2}$
Number of elements $(a,b)$ in the case of $lcm\{p^{2},1\}$ is $ p(p-1) \times 1 =p(p-1)$
So, total number of elements of order $p^2$ is $p(p-1)+p(p-1)^{2}+p^{2}(p-1)^{2}+p(p-1)^{2}+p(p-1)= p^{2}(p^{2}-1)$
According to your lemma, number of cyclic subgroups of order $p^2$ is $\frac{p^{2}(p^{2}-1)}{\phi(p^2)}=p(p+1)$
Clearly the subgroups other than these cyclic subgroups can't contain an element of order $p^2$
So, abelian subgroups other than the cyclic subgroups of order $p^2$ must be of form $A\times B $ where $A , B$ are subgroups of order $p$.
Clearly, that kind of subgroup in $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p^3}$ is only $\mathbb{Z}_{p}\times \mathbb{Z}_{p}$, ( as , there is only one $p$- order subgroup in both $\mathbb{Z}_{p^2}, \mathbb{Z}_{p^3}$)