# Finding the number of subgroups of $\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$.

What is the number of subgroups of $$\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$?

I'm not really sure how to figure it out. I tried seeing subgroups of each $$\Bbb Z_{p^n}$$ but I'm not sure I'm going to get them all.

• The Sylow theorems tell me it exists and that there is only in $\mathbb{Z}_{p^2}$, i dont know how to use them for the other one. A group of order p^2 isnt a Sylow-p subggroup in there. – Pedro Santos Dec 30 '18 at 12:23
• Sylow's theorems will not help you with this question. They do not tell you anything about groups of prime power order. This question is not particularly easy, but you should be able to make a start. For example, can you determine how many subgroups there are of order $p$? – Derek Holt Dec 30 '18 at 12:28
• OK, so if there are $n$ elements of order $p$, then there are $n/(p-1)$ subgroups of order $p$. Also note that all elements of order $p$ are contained in the unique subgroup that is isomorphic to $Z_p \oplus Z_p$. – Derek Holt Dec 30 '18 at 12:47
• Please don't change the question like this. – Shaun Dec 30 '18 at 13:28
• I changed the question back to the original to avoid confusion. – Derek Holt Dec 30 '18 at 14:15

The given group is $$\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$. We will use the formula $$|(m,n)|=lcm(|m|,|n|),$$ for any $$m \in \mathbb{Z}_{p^3}$$ and $$n \in \mathbb{Z}_{p^2}$$, where $$|.|$$ denote the order .

Since $$m,n$$ are the elements of $$\mathbb{Z}_{p^3}$$ and $$\mathbb{Z}_{p^2}$$ respectively, we have $$|m|$$ divides $$p^3$$ and $$|n|$$ divides $$p^2$$.

(i) Show that there is only one subgroup of order of $$\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$ isomorphic to $$\mathbb{Z}_p \oplus \mathbb{Z}_p$$.

In this step we calculate the number of element $$(m,n)$$ of order $$p$$. You easily check that there are $$p^2-1$$ elements having order $$p^2$$. All these $$p^2-1$$ elements along with the identity element $$(0,0)$$ form a unique subgroup of order $$p^2$$ that is isomorphic to $$\mathbb{Z}_p \oplus \mathbb{Z}_p$$.

(ii) Show that there are $$p^2+p$$ subgroup of $$\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$ isomorphic to $$\mathbb{Z}_{p^2}$$.

In this step we will count the number of cyclic subgroups of order $$p^2$$. In other words we have to find all elements $$(m,n)$$ of the group $$\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$ having order $$p^2$$. To do this we need to find

$$(a)$$ all elements $$(m,n) \in \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$, such that $$|m|=p^2 \in \mathbb{Z}_{p^3}$$,

$$(b)$$ all elements $$(m,n) \in \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$, such that $$|n|=p^2 \in \mathbb{Z}_{p^2}$$

From part $$(a)$$ and part $$(b)$$ Try to show that there are total $$p^4-p^2$$ elements of order $$p^2$$ and each cyclic group order $$p^2$$ has $$p^2-p$$ elements of order $$p^2$$.

Hence the number of cyclic subgroup of order $$p^2$$ is equal to $$\frac{p^4-p^2}{p^2-p} =p^2+p$$.

Thus there is a total of $$p^2+p+1$$ subgroups of the group $$\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$$ of order $$p^2$$.

• Alright Thanks, But my problem is that i dont know how to prove those statements. – Pedro Santos Dec 30 '18 at 13:04
• You only seem to be counting the subgroups of order $p^2$. There are others. – Jeremy Rickard Dec 30 '18 at 13:11
• I only wanted those of order p^2, still dont know how to do it though – Pedro Santos Dec 30 '18 at 13:13
• @PedroSantos, I have calculated the result for subgroup of order $p$2$only – M. A. SARKAR Dec 30 '18 at 13:15 • Yes i know thanks, but im not sure how you did it . – Pedro Santos Dec 30 '18 at 13:16 To answer the complete question it is helpful to know that, in a finite abelian group, the number of subgroups of order $$n$$ is equal to the number of index $$n$$ for any $$n$$. This is basically because finite abelian groups $$G$$ are isomorphic to their dual groups $${\rm Hom}(G,{\mathbb C}^\times)$$, and subgroups of the group correspond to quotients of the dual. So the final answer is $$2(1 + (p+1) + (p^2+p+1)) = 2p^2+4p+6$$. • I only knew that finitely generated free modules where isomorphic to their dual, didnt know that a similiar statement worked for finitely generated abelian group and those$Hom\$ – Pedro Santos Dec 30 '18 at 17:09
• That follows from the fact that finite abelian groups are direct sums of cyclic groups. – Derek Holt Dec 30 '18 at 18:07