Question

Show that$\mathit{G}\mathbf{\cong }\mathit{Q}$ under the correspondence
${\mathit{a}}^{\mathbf{r}}{\mathit{b}}^{\mathbf{s}}\mathbf{\to }{\mathit{i}}^{\mathbf{r}}{\mathit{j}}^{\mathbf{s}}\mathbf{(}\mathbf{0}\mathbf{\le }\mathbf{r}\mathbf{\le }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\le }\mathbf{s}\mathbf{\le }\mathbf{1}\mathbf{)}$
by comparing the table in part (a) with the table for Q (see Exercise 16 in Section 7.1).

Short answer

It is shown that$G\cong Q$.

Step-by-step solution

In part (a), we get:

	e	a	a2	a3	b	ab	a2b	a3b
e	e	a	a2	a3	b	ab	a2b	a3b
a	a	a2	a3	e	ab	a2b	a3b	b
a2	a2	a3	e	a	a2b	a3b	b	ab
a3	a3	e	a	a2	a3b	b	ab	a2b
b	b	a3b	a2b	ab	a2	a	e	a3
ab	ab	b	a3b	a2b	a3	a2	a	e
a2b	a2b	ab	b	a3b	e	a3	a2	a
a3b	a3b	a2b	ab	b	a	e	a3	a2

Replace arbs→irjs

	1	i	-1	-i	j	k	-j	-k
1	1	i	-1	-i	j	k	-j	-k
i	i	-1	-i	1	k	-j	-k	j
-1	-1	-i	1	i	-j	-k	j	k
-i	-i	1	i	-1	-k	j	k	-j
j	j	-K	-j	k	-1	i	1	-i
k	k	j	-k	-j	-i	-1	i	1
-j	-j	K	j	-k	1	-i	-1	i
-k	-k	-j	k	j	i	i	-i	-1

Hence,$G\cong Q$.