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I am trying to get a group with 8 elements. This is a Cayley table of $\mathbb Z_2 \times \mathbb Z_4$. Is this right?

$$ \begin{array}{c|cccc} & 1 & 2 & 3 & 4\\ \hline 1 & 1 & 2 & 3 & 0\\ 2 & 2 & 0 & 2 & 0 \end{array} $$

Thanks.

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    $\begingroup$ No, a Cayley table should be square, in your case with 8 elements across the top and left side. $\endgroup$ – Edward Jiang Dec 26 '14 at 21:25
  • $\begingroup$ $\Bbb Z_2\times\Bbb Z_4$ definitely has 8 elements $\endgroup$ – Andrea Mori Dec 26 '14 at 21:33
  • $\begingroup$ jeffreybarr.com/thesis/Documents/BarrThesisWithCode.pdf#page=19 $\endgroup$ – Edward Jiang Dec 26 '14 at 21:35
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    $\begingroup$ You can use the array environment to create a table. Your table isn't correct but I just reformatted it for you. $\endgroup$ – André 3000 Dec 26 '14 at 22:25
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You said you want a group with $8$ elements. Here are your options.

  • Quaternion group $$Q_3\Bigg\{\pm \begin{pmatrix}1 & 0\\0&1\end{pmatrix} ; \pm \begin{pmatrix}i & 0\\0&-i\end{pmatrix} ; \pm \begin{pmatrix}0 & 1\\-1&0\end{pmatrix} ; \pm \begin{pmatrix}0 & i\\i&0\end{pmatrix}\Bigg\}$$

with only one element of order $2$ and it's not abelian;

  • $$\mathbb{Z}/8\mathbb{Z} \{\overline{0},\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7}\}$$ cyclic with an element of order $8$;

  • $$\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} = \{(\overline{0},\overline{0}),(\overline{0},\overline{1}),(\overline{1},\overline{0}),(\overline{1},\overline{1}),(\overline{2},\overline{0}),(\overline{2},\overline{1}),(\overline{3},\overline{0}),(\overline{3},\overline{1})\}$$

  • $$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}=\{(\overline{0},\overline{0},\overline{0}),(\overline{0},\overline{0},\overline{1}),(\overline{0},\overline{1},\overline{0}),(\overline{0},\overline{1},\overline{1}),(\overline{1},\overline{0},\overline{0}),(\overline{1},\overline{0},\overline{1}),(\overline{1},\overline{1},\overline{0}),(\overline{1},\overline{1},\overline{1})\}$$
  • $$D_4 = \{Id, \alpha,\alpha^2,\alpha^3,\beta,\alpha\beta,\alpha^2\beta,\alpha^3\beta\}$$ where $\alpha = \binom{1234}{2341}$ and $\beta= \binom{1234}{4321}$ has $5$ elements of order $2$ it's not abelian.
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  • $\begingroup$ And the way to show they are distinct is by examining subgroups right? $\endgroup$ – SalmonKiller Dec 27 '14 at 1:20
  • $\begingroup$ Not really, I gave some characteristics from each one, so you could see the difference between them, those are all the groups of order $8$, each one with its distinction. For example, $\mathbb{Z}/8\mathbb{Z}$ is abelian and has an element of order $8$, while $D_4$ is not abelian and has one element of order $4$ and $5$ of order $2$. And so it goes... $\endgroup$ – Aaron Maroja Dec 27 '14 at 2:03

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