I am trying to find two distinct group extension of $Z_2$ by $Z_3$. One "natural" extension that I found was $(Z_3,Z_2)$. That is $0 \rightarrow Z_3\rightarrow (Z_3,Z_2)\rightarrow Z_2\rightarrow 0$. However, I am having difficulty in finding another one. Any hints?

  • $\begingroup$ Is $(\Bbb{Z}_3, \Bbb{Z}_2) = \Bbb{Z}_3 \times \Bbb{Z}_2$, the cartesian product? $\endgroup$ – Sammy Black Oct 13 '14 at 6:29
  • $\begingroup$ Do you know about semidirect products? $\endgroup$ – Sammy Black Oct 13 '14 at 6:30
  • $\begingroup$ @SammyBlack: Yes it is cartesian product. Yes, I am familliar with semidirect products $\endgroup$ – Rutherford Mark Oct 13 '14 at 6:36

There are two (split) extensions, depending on which homomorphism you choose $$ \Bbb{Z}_2 \to \operatorname{Aut} \Bbb{Z}_3 \cong C_2. $$ (I prefer the cyclic group notation $C_2$ for automorphisms that are written multiplicatively, although, of course, $C_2 \cong \Bbb{Z}_2$ as abstract groups.)

If you choose the trivial homomorphism, then you get the direct product. Whereas, if you choose the isomorphism, then you get the semidirect product that is isomorphic to $D_3 \cong S_3$, the symmetries of an equilateral triangle. (The first map embeds $\Bbb{Z}_3$ as the subgroup $A_3$ of rotations and the second projects onto the sign of the permutation.)

  • $\begingroup$ Will these two extensions be non isomorphic. It seems like they won't be. $\endgroup$ – Rutherford Mark Oct 13 '14 at 6:50
  • $\begingroup$ $\Bbb{Z}_3 \times \Bbb{Z}_2 \cong \Bbb{Z}_6$ is abelian, while $\Bbb{Z}_3 \rtimes \Bbb{Z}_2 \cong S_3$ is not. $\endgroup$ – Sammy Black Oct 13 '14 at 6:55
  • 1
    $\begingroup$ There are not so many groups of order six anyway $\endgroup$ – Hagen von Eitzen Oct 13 '14 at 9:14

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