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How many subgroups of order $p^2$ does the abelian group $\mathbb{Z_{p^3}} \times \mathbb{Z_{p^2}}$have ?

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First step in answering is to ask how many non-isomorphic groups of order $p^2$ are there, and what is their structure? – Gerry Myerson May 4 '14 at 7:51
It will either be isomorphic to $\mathbb{Z_{p^2}}$ or be isomorphic to $\mathbb{Z_p}$ x $\mathbb{Z_p}$ – user119065 May 4 '14 at 8:46
Good. That should help you to find the number of instances of each of those as subgroups of your group. – Gerry Myerson May 4 '14 at 9:13
This question is not as trivial as it seems, since a subgroup of $A\times B$ need not to be in the form of $H\times K$. As an easy example, $Z_p\times Z_p$ has $p+1$ subgroup isomorphic to $Z_p$.(not 2). – mesel May 4 '14 at 10:25
Yes, that is the problem. Your example is not that difficult to work out. Its $p^2$ - $1$ non identity elements partitioned into $p$ - $1$ parts. – user119065 May 4 '14 at 10:46
up vote 0 down vote accepted

lemma: if there are $n$ elements of order $d$ in $G$, there are $\dfrac n {\phi(d)}$ cyclic subgroup of order $d$ where $\phi$ is Euler phi function.

By using this lemma and counting the elements of $G$ with order $p^2$, I found $p^2+p$ cyclic subgroup of $G$. The methhod is simple but calculation is boring.

Here is the calculation.

Let $G=Z_{p^2}\times Z_{p^3}$ and $a$ be an element of $G$ with order $p^2$ then $a=(x,y)$ where $lcd(|x||,|y|=p^2)$ then at least one of the order is $p^2$.

case 1: $|x|=p^2$

$\phi(p^2)=p^2-p$ if $|y|=p$ we have $p-1$ such $y$ as a result we get $(p-1)(p^2-p)$ elements in $G$.

if $|y|=p^2$ then we get $(p^2-p)$ such $y$ as a result $(p^2-p)^2$ elements in $G$.

$if |y|=1$ then we mest have $y=e$ so $(p^2-p)$ elements in $G$.

In total, we have $p^2(p^2-p)$ elements in $G$.

case 2: if $|y|=p^2$ then it is enough to calculate when $|x|=p$ and $|x|=e$ .(we already calculated the case $|x|=|y|=p^2$).

Wits same manner we get $p(p^2-p)$ elements in $G$.

From case $1$ and case $2$ we get $(p^2+p)(p^2-p)$ elements of order $p^2$ in $G$.

Thus the number of the cyclic subgroup of $G$ with order $p^2$ is $\dfrac {(p^2+p)(p^2-p)}{\phi(p^2)}=p^2+p$

Edit: Notice that the group $\mathbb Z_p \times Z_p$ has $p^2-1$ elements of order $p$.

$\mathbb Z_{p^2}$ has $p-1$ elements of order $p$ and $Z_{p^3}$ has $p-1$ elemets of order $p$ . and $G$ have $(p-1)p$+$1.(p-1)=p^2-1$ elements of order $p$. So $G$ has uniqe subgroup which is isomorphic to $\mathbb Z_p \times \mathbb Z_p $.

In total: $G$ has $p^2+p+1$ subgroup of order $p^2$.

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I will edit my answer to comlete the answer. – mesel May 4 '14 at 11:09
@user119065: I hope the lemma used in the answer is clear for you. – mesel May 4 '14 at 11:26
Can you explain how the lemma is working ? – user119065 May 4 '14 at 12:32
A cyclic group with order $d$ has $\phi(d)$ generaters and of course all of them has order $d$. If there are $k$ cyclic subgroup of order $d$ in $G$ then there are $k\phi(d)$ elements of order $d$ in $G$. (Notice that even if the cyclic subgroups may intersects, generators will not be in intersection. This lemma does not even require $G$ is abelian ). Example: In $S_4$, there are $(4-1)!=6$ elements of order $4$ so $\dfrac 6 {\phi(4)}=3$ cyclic subgroup of order $4$. – mesel May 4 '14 at 12:42

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