In $Q_8$ we have the following elements $$ \{ 1,-1,i,-i,j,-j,k,-k \} $$ Here $|1| = 1$. We have that $|-1| = 2$. We have that $|i| = |j| = |k| = 4$. Now $$ (-i)^2 = -1 \cdot i \cdot -1 \cdot i = i \cdot -1 \cdot -1 \cdot i = i^2 = -1 $$ So $|-i| = |-j| = |-k| = 4$. In total
1 | -1 | i | -i | j | -j | k | -k |
1 | 2 | 4 | 4 | 4 | 4 | 4 | 4 |
For the multiplication table we get
x | 1 | -1 | i | -i | j | -j | k | -k |
1 | 1 | -1 | i | -i | j | -j | k | -k |
-1 | -1 | 1 | -i | i | -j | j | -k | k |
i | i | -i | -1 | 1 | k | -k | -j | j |
-i | -i | i | 1 | -1 | -k | k | j | -j |
j | j | -j | -k | k | -1 | 1 | i | -i |
-j | -j | j | k | -k | 1 | -1 | -i | i |
k | k | -k | j | -j | -i | i | -1 | 1 |
-k | -k | k | -j | j | i | -i | 1 | -1 |
The group can be generated with two of the three letters. For example $i,j$. Here we need four of the relations stated in the book, that is
In total we get $$ \langle i,j\ |\ R_1 , R_2 , R_3 , R_4 \rangle $$