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Let $G$ be a group of order $21$. Find the number of non-isomorphic groups of order $21$

My solution: If the group $G$ is commutative,then $G$ can be expressed as a direct product of cyclic groups of prime power order,i.e it is either $\mathbb Z_3\times \mathbb Z_7$ or $\mathbb Z_{21}$ which is same

My problem is when $G$ is commutative.How to proceed here?

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  • $\begingroup$ Do you know something about group of order pq? $\endgroup$
    – Arpit Kansal
    Mar 14, 2015 at 11:28
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    $\begingroup$ This is a rather general hint: use Sylow theorems. $\endgroup$
    – Eclipse Sun
    Mar 14, 2015 at 12:06

1 Answer 1

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Let $p>q$ be primes. There exists a non-abelian group of order $pq$ if and only if $p\equiv 1 \bmod q$. Furthermore any tow non-abelian groups of order $pq$ are isomorphic. For a proof use the Sylow theorems. Also, one can find this result in many books. For $p=7$ and $q=3$ we obtain that there is exactly one non-abelian group of order $21$. The abelian groups are $C_{21}\simeq C_3\times C_7$.

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  • $\begingroup$ why any two non-abelian groups are isomorphic?in which books I can find this $\endgroup$
    – Learnmore
    Mar 15, 2015 at 2:26
  • $\begingroup$ In a text on Sylow theorems, e.g., Theorem $4.5$ here. $\endgroup$
    – Dietrich Burde
    Mar 15, 2015 at 11:50

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