# Did I map this group right?

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.

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

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.
• And the way to show they are distinct is by examining subgroups right? – SalmonKiller Dec 27 '14 at 1:20
• 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... – Aaron Maroja Dec 27 '14 at 2:03