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What is the smallest example of a group that is not isomorphic to a cyclic group, a direct product of cyclic groups or a semi-direct product of cyclic groups?

So finite abelian groups are ruled out, as are the groups up to order $7$ I believe, since $1,2,3$ are all isomorphic to cyclic groups, $4$ has two non-isomorphic groups one cyclic and one isomorphic to the direct product $\Bbb{Z/2Z\times Z/2Z}$. $5,7$ are prime and so must be cyclic, and there's only one nonabelian group of order $6$ which is isomorphic to $\Bbb{Z/3Z\rtimes Z/2Z}$. Is it the quaternion group?

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4 Answers 4

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The quaternion group is indeed the minimal counter-example. Clearly, any group of orders $1,2,3,5$ or $7$ is cyclic, and the two (non-isomorphic) groups of order $4$ are:

$\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$.

There are likewise just two non-isomorphic groups of order $6$:

$\Bbb Z_6$ and $S_3 \cong \Bbb Z_3 \rtimes \Bbb Z_2$ (this is the only possible non-trivial semi-direct product of these two groups, since:

$0 \mapsto 1_{\Bbb Z_3}\\1 \mapsto (x \mapsto -x)$

is the sole non-trivial homomorphism $\Bbb Z_2 \to \text{Aut}(\Bbb Z_3)$).

It is convenient to use this formulation of a(n internal) semi-direct product:

  1. $G = NH$, where $N,H$ are subgroups of $G$, and $N \lhd G$.

  2. $N \cap H = \{e_G\}$.

The problem with obtaining $Q_8$ as a semi-direct product of two proper subgroups, is that we must have either $|H|$ or $|N|$ equal to $2$. But the only subgroup of order $2$ in $Q_8$ is $\{1,-1\}$, which is a subgroup of every subgroup of $Q_8$ of order $4$:

$\langle i\rangle = \{1,-1,i,-i\}\\\langle j\rangle = \{1,-1,j,-j\}\\\langle k\rangle = \{1,-1,k,-k\}$

Note that all $6$ elements of order $4$ lie in one of these $3$ subgroups.

Thus the condition $N \cap H = \{e_G\}$ (which is $=\{1\}$ in this case) is impossible to satisfy.

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    $\begingroup$ I'm accepting your answer, because it is the most complete and very clearly presented. Thank you! $\endgroup$
    – snulty
    Aug 28, 2015 at 9:09
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Yes, quaternion group is non abelian and not isomorphic to the direct product of $\Bbb{Z/2Z\times Z/2Z\times Z/2Z}$ or $\Bbb{Z/2Z\times Z/4Z}$, and is the one you are looking for.

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    $\begingroup$ But we need a stronger result, namely that it is not even isomorphic to a semidirect product. $\endgroup$ Aug 27, 2015 at 21:29
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    $\begingroup$ See here: math.stackexchange.com/questions/1054242/… $\endgroup$ Aug 27, 2015 at 21:34
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    $\begingroup$ @LorenoHeer thanks for the link, leppala's answer there is very clean and clear! $\endgroup$
    – snulty
    Aug 27, 2015 at 23:05
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Not sure but a good candidate should be the quaternion group. If not, the alternate group $A_5$ (much bigger), as it is a simple group.

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  • $\begingroup$ I like the offered alternative of a simple group! $\endgroup$
    – snulty
    Aug 27, 2015 at 23:02
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Although this question is 6 years old, I have another example. It is a group G of order 48 that has the short exact sequence $1\mapsto SL(2,3) \mapsto G \mapsto C_2 \mapsto 1$, called CSU(2,3) or G[48,28] in the Groupnames site. This group has 3 subgroups isomorphic to $Q_{16}$ and 4 subgroups isomorphic to $C_3$, and so none of its Sylow subgroups is normal. Therefore if it is a semi-direct product, it must be $H \rtimes K$ where $K$ is normal, so that both $H$ and $K$ have even order, and are disjoint. By Lagrange's Theorem, both $H$ and $K$ must have an element of order 2, and these two elements have to be different. But CSU(2,3) has only one element of order 2, so it is not a semi-direct product.

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