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Given the abelian groups of order $7425$:

$$Z_{33} \times Z_{15} \times Z_{15} , \ Z_{25} \times Z_{297} , \ Z_{45} \times Z_{165} , Z_{55}\times Z_9 \times Z_{15}$$

  • Which of these groups, if any, are isomorphic?

  • Which groups are cyclic?

  • How can one write these group as a product of cyclic groups of prime-power order.


I found that $Z_{25} \times Z_{297}$ is cyclic while the others are not. I'm not sure which are isomorphic.

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  • $\begingroup$ I found Z25×Z297 cyclic while others not .I'm not sure which isomorphic $\endgroup$ – M Alrantisi Dec 12 '14 at 13:56
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I assume you are familiar with this lemma (or can prove it): If $n$ and $m$ are coprime, then $Z_{nm}=Z_n\times Z_m$ (where equality stands for isomorphism).

Using this lemma you can decompose all of your groups to products of cyclic groups of prime power order. For example, \begin{eqnarray} Z_{33}\times Z_{15}\times Z_{15} &=& Z_{3\cdot11}\times Z_{3\cdot5}\times Z_{3\cdot5} \\&=& Z_3\times Z_{11}\times Z_3\times Z_5\times Z_3\times Z_5. \end{eqnarray} From here you can recognize that this group has a noncyclic subgroup $Z_3\times Z_3$, so the group itself is not cyclic.

A product of cyclic groups of coprime orders is cyclic, as is easy to check. With these tools you can answer question 2.

To answer question 1, you can use a converse of the lemma I stated at the beginning. But there is also a more direct way now that you have explicit groups. If two groups have the same decomposition (as in your last question), they are isomorphic. But for example $Z_5\times Z_5\neq Z_{25}$ because only one of these groups has an element of order 25.

Can you solve your problem and check your results with these ideas?


Addendum: With the method I described you get \begin{eqnarray} Z_{33}\times Z_{15}\times Z_{15} &=& Z_3\times Z_{11}\times Z_3\times Z_5\times Z_3\times Z_5,\\ Z_{25}\times Z_{297} &=& Z_{25}\times Z_{27}\times Z_{11},\\ Z_{45}\times Z_{165} &=& Z_5\times Z_9\times Z_3\times Z_5\times Z_{11}\quad\text{and}\\ Z_{55}\times Z_9\times Z_{15} &=& Z_5\times Z_{11}\times Z_9\times Z_3\times Z_5. \end{eqnarray} Using the ideas I gave above, one can see that the second one is the only cyclic group, so others are not isomorphic to it. Compare the decompositions of the last two groups: What do you notice? Compare the first and the third group: Which one of these has elements of order 9?

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  • $\begingroup$ Thanks but I still can't get isomorphism can you help on this please.I found all not cyclic except z25×Z297. $\endgroup$ – M Alrantisi Dec 12 '14 at 14:41
  • $\begingroup$ @MAlrantisi, I added something to help with the isomorphism problem. You should be able to take it from here. $\endgroup$ – Joonas Ilmavirta Dec 12 '14 at 14:59
  • $\begingroup$ Thank joonas .it's a great help I found that first and third are isomorphic also 3 and 4 are isomorphic depending on same structure decomposing $\endgroup$ – M Alrantisi Dec 12 '14 at 15:38
  • $\begingroup$ @MAlrantisi, the third and fourth group are indeed isomorphic, but the first one is not isomorphic to them. (Look at the hint regarding elements of order 9. There is a difference in this property, so the groups are not isomorphic.) Anyway, I was glad to be of service! $\endgroup$ – Joonas Ilmavirta Dec 12 '14 at 20:40
  • $\begingroup$ Thanks its been a fabulous service you gave.and I understand it now $\endgroup$ – M Alrantisi Dec 12 '14 at 21:27

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