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. 
 A: 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. 

