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I'm interested in understand the classification of finite groups of order less then 10. We have of course that for the Lagrange Theorem all groups of prime order such as $2,3,5,7$ are cyclic and therefore isomorphic to $\mathbb Z_2, \mathbb Z_3, \mathbb Z_5,\mathbb Z_7$. Then we have groups that have order $p^2$ such as $4$ and $9$ which have to be abelian and therefore isomorphic to $\mathbb Z_4, \mathbb Z_2 \times\mathbb Z_2$ and $\mathbb Z_9, \mathbb Z_3 \times\mathbb Z_3$. Now we have order 6 and 8. I'd like to make more precise the following statement: every group of this kind is a direct or semidirect product of $\mathbb Z_2$ with a normal subgroup, i.e.

order 6: can be if abelian $\mathbb Z_3 \times \mathbb Z_2 $ or if not abelian can only be $\mathbb Z_3 \rtimes \mathbb Z_2$

order 8: can be if abelian $\mathbb Z_8$ if has an element of order 8, $\mathbb Z_4 \times \mathbb Z_2 $ if doesn't have an element of order 8 but has one of order 4, $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 $ if it has only elements of order 2. If it is not abelian can be only or if not abelian can only be $\mathbb Z_4 \rtimes \mathbb Z_2 $ if has one element of order 4, $\mathbb Z_2 \times \mathbb Z_2 \rtimes \mathbb Z_2 $.

Is this true? How can I be more precise showing the automorphism of the semidirect product? How can I effectively demonstrate the part relatively of the order 8 groups? Is this way of proceeding generalizable?

Thank you in advance

Olexandr Konovalov
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Dac0
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  • Your list of groups of order 8 is missing the quaternion group. Running the risk of blowing my own trumpet you can take a peek at this. Also, a group with all the elements of order two is necessarily abelian. Proof is a standard exercise. – Jyrki Lahtonen May 19 '17 at 07:47
  • Mind you, there is a non-trivial (and non-abelian) semidirect product $(\Bbb{Z}_2\times\Bbb{Z}_2)\rtimes\Bbb{Z}_2$, but that turns out to be isomorphic to the dihedral group. For example you can think of that normal subgroup isomorphic to Klein four as the group generated by reflections w.r.t. the coordinate axes. Then the extra $\Bbb{Z}_2$ comes from reflecting w.r.t. the line $y=x$. – Jyrki Lahtonen May 19 '17 at 07:52
  • It's true...I've also done this exercise :D :D :D, Anyway does it still stand the $\mathbb Z_2 \times \mathbb Z_2 \rtimes \mathbb Z_2 $ assumption? This should be the quaternion group as $V \rtimes \mathbb Z_2 $ or I'm doing something wrong? It has been a long time since I've studied these topics – Dac0 May 19 '17 at 07:53
  • Can I write the Quaternion group as a semidirect product of something? – Dac0 May 19 '17 at 07:54
  • No. The quaternion group cannot be written as a semidirect product of smaller groups. That's basically because it has a single element of order two, and necessarily the two component groups would have at least one each. – Jyrki Lahtonen May 19 '17 at 07:54
  • perfectly clear! thank you – Dac0 May 19 '17 at 07:55

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