Question

Show that T is a nonabelian subgroup of G.

Short answer

It is proved that T is a non-abelian subgroup of$G$.

Step-by-step solution

Step-by-Step Solution Step 1: Referring to Theorem 7.12

Theorem 7.12

Let H be a non-empty finite subset of group G. If H is closed under the operation in G, then H is a subgroup of G.

Proving that T is a non-abelian subgroup of G

Referring to part (a) of this exercise, we know the following:

$\begin{array}{c}{\mathrm{a}}^{\mathrm{i}}{\mathrm{a}}^{\mathrm{j}}={\mathrm{a}}^{\mathrm{i}+\mathrm{j}\mathrm{mod}6}\in \mathrm{T}\\ {\mathrm{a}}^{\mathrm{i}}{\mathrm{a}}^{\mathrm{j}}\mathrm{b}={\mathrm{a}}^{\mathrm{i}+\mathrm{j}\mathrm{mod}6}\mathrm{b}\in \mathrm{T}\\ {\mathrm{a}}^{\mathrm{i}}\mathrm{b}{\mathrm{a}}^{\mathrm{j}}={\mathrm{a}}^{\mathrm{i}-\mathrm{j}\mathrm{mod}6}\mathrm{b}\in \mathrm{T}\\ {\mathrm{a}}^{\mathrm{i}}\mathrm{b}{\mathrm{a}}^{\mathrm{j}}\mathrm{b}={\mathrm{a}}^{\mathrm{i}-\mathrm{j}+3\mathrm{mod}6}\in \mathrm{T}\end{array}$

From the relation shown above, we conclude that T is closed under the composition.

And for any arbitrary,${\mathrm{a}}^{\mathrm{i}}\in \mathrm{T},$we know that ${\left({\mathrm{a}}^{\mathrm{i}}\right)}^{-1}={\mathrm{a}}^{6-\mathrm{i}}\in \mathrm{T}$and ${\left({\mathrm{a}}^{\mathrm{i}}\mathrm{b}\right)}^{-1}={\mathrm{a}}^{\mathrm{i}-3\mathrm{mod}6}\mathrm{b}\in \mathrm{T}$.

Therefore, T is also closed under the inverse composition.

Hence, fromTheorem 7.12, we can conclude that T is a non-abelian subgroup of G.