# Is this object a group?

$\begin{array}{ccccccccc} \times&e_0&e_1&e_2&e_3&e_4&e_5&e_6&e_7\\ e_0&I&e_1&e_2&e_3&e_4&e_5&e_6&e_7\\ e_1&e_1&-I&e_3&-e_2&e_5&-e_4&e_7&-e_6\\ e_2&e_2&e_3&-I&-e_1&e_6&e_7&-e_4&-e_5\\ e_3&e_3&-e_2&-e_1&I&e_7&-e_6&-e_5&e_4\\ e_4&e_4&e_5&-e_6&-e_7&-I&-e_1&e_2&e_3\\ e_5&e_5&-e_4&-e_7&e_6&-e_1&I&e_3&-e_2\\ e_6&e_6&e_7&e_4&e_5&-e_2&-e_3&-I&-e_1\\ e_7&e_7&-e_6&e_5&-e_4&-e_3&e_2&-e_1&I \end{array}$

There is an identity element and each element has an inverse but is the multiplication associative? To check manually would require $2\times8^3$ multiplications. I'm hoping there is a more elegant way than by brute force. If this is a group, is it $Q_8$?

If it helps, $e_0,e_2,e_4,e_6$ are quaternions.

• What's with the minus signs? Is no such thing in a (multiplicatively written) group. – Henning Makholm Apr 8 '14 at 1:05
• If the minuses mean that you actually have $16$ elements, $e_0$, $e_1$, ..., $e_7$, $-e_0$, $-e_1$, ..., $-e_7$, then there are too many for them to be the $Q_8$. And they're not octonions either, because there are too many elements of order 2. – Henning Makholm Apr 8 '14 at 1:13
• This is an incomplete table as the people above me have pointed out. Furthermore, should you replace $I$ with $e_0$? Or does $I$ not denote the identity? I recommend you give us the background if you want your question answered. How did you build this table? Is it given geometrically, by a system of equations, etc? – JMag Apr 8 '14 at 3:56

It is the group $(C_4\times Q_8)/\{(0,0),(2,-1)\}$ with $$\begin{array}{c} e_0 = (0,1) \qquad e_2 = (0,i) \qquad e_4 = (0,j) \qquad e_6 = (0,k) \\ e_1 = (1,1) \qquad e_3 = (1,i) \qquad e_5 = (1,j) \qquad e_7 = (1,k) \end{array}$$
This can also be presented as the group generated by $\{e_1,e_2,e_4\}$ with the relations $$e_1^4=1 \qquad e_1^2=e_2^2=e_4^2 \qquad e_1e_2=e_2e_1 \qquad e_1e_4=e_4e_1 \qquad e_4e_2 = e_1^2e_2e_4$$
Then, in your table, $e_0=1$ and $e_6=e_2e_4$ and $e_{2n+1} = e_1e_{2n}$ and $-e_n = e_2^2e_n$.