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For example, let's say we have a group Q8 (Quaternion group).

How to prove that this group is not isomorphic to any member of any of the families of groups (Cyclic, Abelian, Dihedral, Symmetric, Alternating)?

What is the general approach for solving this problem for any given group?

Oleg Dats
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  • For example, you know the order of $Q_8$ and hence only have very few candidates in your families of groups. Of course, for abelian, you just have to exhibit two non-commuting elements. – Hagen von Eitzen Jun 04 '20 at 16:00
  • Given two presentations, the task, in general, of deciding whether they define isomorphic groups is undecidable. This is known as the isomorphism problem – Shaun Jun 04 '20 at 16:15

2 Answers2

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It is clear that $Q_8$ is not abelian, hence not cyclic, because for example $ij=-ji$. So we can use, if we want, that we only have two nonabelian groups of order $8$, namely $Q_8$ and $D_4$. So we only need to distinguish these two groups. But this is easy. Every subgroup in $Q_8$ is normal, which is not true for $D_4$.

References:

How can you show there are only 2 nonabelian groups of order 8?

Show that every subgroup of $Q_8$ is normal.

Dietrich Burde
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It's not cyclic, since there are no elements of order $8$. Neither is it abelian, since $ij \neq ji$. It can't be Dihedral, as $D_8$has four elements of order $2$ whereas the quarternions only have one. It is neither symmetric nor alternating, since these groups have sizes of $n!$ and $\frac{n!}{2}$ respectively, and $8$ is not of these forms.

hdighfan
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