Question

Show that T is not isomorphic to ${\mathit{D}}_{\mathbf{6}}$ or to ${\mathit{A}}_{\mathbf{4}}$.

Short answer

It is proved that T is not isomorphic to${\mathrm{D}}_6$ or to${\mathrm{A}}_4$ .

Step-by-step solution

Step-by-Step Solution Step 1: Definition of Isomorphism

Let G and H be groups with the group operation denoted by$\mathbf{\ast }$. G is isomorphic to a group H (in symbols$\mathit{G}\mathbf{\cong }\mathit{H}$ ), if there is a function$\mathbf{f}\mathbf{:}\mathbf{G}\mathbf{\to }\mathbf{H}$, such that:

$\mathbf{1}\mathbf{)}\mathit{f}\mathbf{(}\mathbf{a}\mathbf{\ast }\mathbf{b}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\ast }\mathit{f}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$

2) fis injective.

3)fis surjective.

Showing that T is not isomorphic to D6

$D_6$is a dihedral group of degree 3 and order 6, and it has six elements of order 2.

Since T has only one role="math" localid="1652865462340" ${\mathrm{a}}^3$element, therefore, there cannot be a function role="math" localid="1652865641002" $f:T\to D_6$, which satisfies the condition for Isomorphism.

Hence, T is not isomorphic to $D_6$.

Showing that T is not isomorphic to A4

The alternating group $A_4$is acyclic group and it has 3 elements of order 2.

Since T has only one role="math" localid="1652865614026" ${\mathrm{a}}^3$element, therefore, there cannot be a function $f:T\to A_4$, which satisfies the condition for isomorphism.

Hence, T is not Isomorphic to $A_4$.