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.

My answer:

$p^2, q^2, q$

$p,p, q^2, q$

Is this correct?


No it's not correct.


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

  • $\begingroup$ Why not the other two groups? It is because of the $q^2$ containing element? $\endgroup$
    – Daniel
    Jan 4 '21 at 11:54
  • 1
    $\begingroup$ Because there are no element of order $q^2$ in $\mathbb Z_q\times \mathbb Z_q\times \mathbb Z_q$ @Daniel $\endgroup$
    – Surb
    Jan 4 '21 at 11:57
  • 1
    $\begingroup$ Possibly, sorry :-) $\endgroup$ Jan 4 '21 at 13:10
  • 1
    $\begingroup$ Yes, I got that wrong too I am afraid. I withdraw my comment that the answer is not correct! $\endgroup$
    – Derek Holt
    Jan 4 '21 at 13:13
  • $\begingroup$ How do you show that there is an element $q^2$ in $\mathbb{Z}_{q^3}$? Is it because $q^3 q^3=q^2 q^2 q^2$? $\endgroup$
    – Daniel
    Jan 21 '21 at 14:54

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