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We need to find the smallest composite integer $n$ such that there is a unique group of order $n$.

Attempt: Let us suppose that $n= ab$ is a composite integer where $a,b$ are integers such that $a \neq 1 , b \neq 1$.

Then, there are always at least two ways to write $G$ whose order $n$ of the form $ab$. For Ex: $G$ can be either $\mathbb Z_{ab}$ or $ \mathbb Z_a \oplus \mathbb Z_b$

Hence, the uniqueness property upto isomorphism for a composite $n$ should never exist. Where could I have gone wrong?

EDIT: I haven't taken the situation when $a,b$ are relatively prime into account.

MathMan
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    If $a$ and $b$ are coprime then $Z_{ab}=Z_a\oplus Z_b$. – Seth Aug 25 '14 at 22:30
  • Hint: Look at $15$. – André Nicolas Aug 25 '14 at 22:32
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    $a$ would have to be a product of distint primes or there are more than 1 abelian group. Then the question is equivalent to asking for $a$ a product of distinct primes when is every group of order $a$ abelian? – Seth Aug 25 '14 at 22:32
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    @AndréNicolas and Seth : As per Seth's argument, $\mathbb Z_2 \oplus \mathbb Z_3 = \mathbb Z_6$ should be the required answer? – MathMan Aug 25 '14 at 22:35
  • oops no because then we have $D_3$ right? Is this search mainly a trial and error one? – MathMan Aug 25 '14 at 22:37
  • Look at symmetric group $S_3$. (Added: In the meantime you posted the comment $D_3$, same). – André Nicolas Aug 25 '14 at 22:39
  • yes, so is it more of a trial and error method? for a group order $n$ I will have to find out if other groups exist or not? – MathMan Aug 25 '14 at 22:41
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    Not really trial and error if you have covered Sylow theorems. The prime $3$ does not divide $5-1$. You do need to rule out the evens $\ge 4$ (and less than $15$) but this is not hard. – André Nicolas Aug 25 '14 at 22:43
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    Showing every group of order $15$ is cyclic is done as an application in the Sylow Theorems article in Wikipedia. – André Nicolas Aug 25 '14 at 22:52
  • All the evens $b\geq 4$ can be written in the form $2 a$ such that $a \neq 1$. Hence, $\mathbb Z_b$ and $\mathbb Z_2 \oplus \mathbb Z_a$ or $S_m$ , $D_m$ are available options. We are left with $9$ and $15$. Now, if $|H|=9, H \approx \mathbb Z_9 $ or $\approx \mathbb Z_3 \oplus \mathbb Z_3$.

    $ H $ when $|H| =15$ is cyclic group as $3$ does not divide $5-1$. and there does not appear to be any dihedral group or permutation group either isomorphic to it . However, how can we be sure that there is no other group at all isomorphic to it?

     – MathMan Aug 25 '14 at 22:53
  • 15 is the first composite cyclic number. See https://en.wikipedia.org/wiki/Cyclic_number_(group_theory) – lhf Nov 08 '23 at 14:57

2 Answers2

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The comments are enough to rule out all composites less than $15$. For example, for even composites we can use the dihedral group and the cyclic group.

To show that every group of order $15$ is cyclic, look at the Sylow Theorems article in wikipedia, where the problem is discussed explicitly. Or else look at this old MSE question.

Community
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André Nicolas
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The group of order p^2 where p is prime i.e group of order 4 and 9 can be eliminated from the fact that these are abelian and so isomorphic to either Zp^2 or Zp×Zp The groups of order 6,8,10,12 can be eliminated because they will be isomorphic to dihedral group and cyclic group So the group of order 15 is cyclic,abelian and hence unique

Nivedita Sharma
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