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I was thinking of building on top of known non Abelian groups, like $S_3$, and taking a direct product with $\Bbb Z_n$'s but those groups' order would be a multiple of order of $S_3$.

  • So, is there is a clever way to do it for any order, like use an Abelian group of order close to, say $39$, and make it non-Abelian?
  • Here, Finding presentation of group of order 39 they give a general representation but can we come up with an actual example without using Sylow theorems? Appreciate your response.
Chinnapparaj R
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    The answer by pasco to the linked question provides an actual example of a nonabelian group of order $39$, namely the group $\langle x,y \mid x^{13},y^3,y^{-1}xy=x^3 \rangle$. This type of construction is known as a semidirect product. – Derek Holt Dec 02 '18 at 11:43
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    If $|G|=pq$ with primes $p\neq q$ with $p\mid q-1$, then there is always a non-abelian group of order $pq$, which is a semidirect product as above, see here, or here. We have $39=pq=3\cdot 13$ and $3\mid 12$. – Dietrich Burde Dec 02 '18 at 12:16
  • Thanks, so semi-direct product of Z13 and Z3 is an example then? –  Dec 02 '18 at 19:22

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There are composite orders (e.g. 15 or 765, or prime squares) such that all groups of that order will be abelian, and there is no all-purpose construction, but here are a few constructions of nonabelian groups that cover lots of orders:

  • If the order is even and $>4$, one can construct a dihedral group.

  • If the order involves a prime power $p^k$ such that another prime divisor (that could be $p$ as well, so this in particular covers all orders that are multiples of a prime cubed) that divides the order of $GL_k(p)$, one can form a nonabelian semidirect product.

ahulpke
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