# A is finite abelian with order $p^2q^3$ and with an element of order $q^2$. p and q are distinct primes. [closed]

In my practice problem for the exam I have:

Suppose $$A$$ is a finite abelian with order $$p^2q^3$$ and with an element of order $$q^2$$, where $$p$$ and $$q$$ are distinct primes.

Show all possible elementary divisors.

$$p^2, q^2, q$$

$$p,p, q^2, q$$

Is this correct?

No it's not correct.

Hint

Notice that $$A$$ is isomorphic to one of the following group :

$$\mathbb Z_{q^2}\times \mathbb Z_q\times \mathbb Z_{p^2}$$ $$\mathbb Z_{q^2}\times \mathbb Z_q\times \mathbb Z_{p}\times \mathbb Z_p$$ $$\mathbb Z_{q^3}\times \mathbb Z_{p^2}$$ $$\mathbb Z_{q^3}\times \mathbb Z_{p}\times \mathbb Z_p$$

where $$\mathbb Z_r:=\mathbb Z/r\mathbb Z$$.

• Why not the other two groups? It is because of the $q^2$ containing element? Jan 4, 2021 at 11:54
• Because there are no element of order $q^2$ in $\mathbb Z_q\times \mathbb Z_q\times \mathbb Z_q$ @Daniel
– Surb
Jan 4, 2021 at 11:57
• Possibly, sorry :-) Jan 4, 2021 at 13:10
• Yes, I got that wrong too I am afraid. I withdraw my comment that the answer is not correct! Jan 4, 2021 at 13:13
• How do you show that there is an element $q^2$ in $\mathbb{Z}_{q^3}$? Is it because $q^3 q^3=q^2 q^2 q^2$? Jan 21, 2021 at 14:54