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If G is a group not cyclic then its order can be:

a)15

b)35

c)77

d)120

e)2011

Well, i know that if G is not cyclic then it is not isomorphic to Zn, but i think it does not help much.

Any tips?

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    $\begingroup$ Hint: Do you know the Sylow Theorems? Or any basic non-cyclic or non-abelian groups? Is this a question where you know there is just one answer, or could there be more than one correct answer? $\endgroup$ May 4, 2015 at 13:32
  • $\begingroup$ Yes, i do. I know the basic of groups, and i have seen the sylow theorems. $\endgroup$ May 4, 2015 at 13:36
  • $\begingroup$ The question is just that, there is no more information. $\endgroup$ May 4, 2015 at 13:37
  • $\begingroup$ Although this comment goes beyond the assumed background of the question, there is a complete classification of the $n$ such that all groups of order $n$ are cyclic: $(n,\varphi(n)) = 1$. Therefore a positive integer is the size of some non-cyclic group if and only if the number does not have that property. Now in your case the situation is much simpler, since there's an obvious non-cyclic group of order 120. $\endgroup$
    – KCd
    May 4, 2015 at 13:58

4 Answers 4

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Proposition Let $n$ be a positive integer. Then there is only one group of order $n$ if and only if gcd$(n,\varphi(n))=1$.

Note that such a group must be necessarily cyclic.

Except for 120 all other numbers satisfy this criterion. $S_5$ has order $120$ and is not cyclic.

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Let $G$ be a group from order $pq$ which $p,q$ are distinct prime numbers and $p\lt q$ and more $q-1$ not divisible to $p$ then, $G$ is cyclic group. And we Know every group from order of a prime number, is cyclic. So, the only possible case is $120$.

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Actually the term G is non-cyclic helps pretty much.It says that G is not isomorphic to Z/n.But one knows that $$ \mathbb Z_{pq}\cong\mathbb Z_p\times \mathbb Z_q, $$when p and q are coprime.Using this arguement you can easily reach at your ans which should be option D.

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  • $\begingroup$ Your argument is not sufficient to reach the conclusion. $\endgroup$ May 4, 2015 at 13:47
  • $\begingroup$ Actually in all the cases except D, the order of the group can be decomposable into mutually prime numbers.So they all are isomorphic to Z/p * Z/q , which is cyclic. $\endgroup$ May 4, 2015 at 13:51
  • $\begingroup$ No, that is incorrect. Not all groups of order $pq$ where $p$ and $q$ are coprime are cyclic. They happen to be here, but you need more. $\endgroup$ May 4, 2015 at 13:52
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If this is a multiple choice question with one answer, then you can find it easily.

There is a nonabelian dihedral group of order $2n$ for ever $n\ge 3$

Or you can notice that if $p^2$ is a factor of the order of your group, then you can create a group of requisite order as a direct product including the non-cyclic component $\mathbb Z_p \times \mathbb Z_p$ (e.g. the non-cyclic group of order $4$)

Or you could know something about the symmetric group $S_5$ of order $120$

Any way, you know one answer. Are the others possible? Well the Sylow theorems deal with the first three (but there is a non-cyclic group of order $21$ because $7\equiv 1 \bmod 3$). And $2011$ is prime.

Moral: you can't really do group theory without having a good understanding of some standard and basic examples of groups.

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