# Thread: Non-abelian subgroup T of $S_{3} \times Z_{4}$

1. ## Non-abelian subgroup T of $S_{3} \times Z_{4}$

There is a nonabelian subgroup T of $\displaystyle S_{3} \times Z_{4}$ of order 12 generated by elements a, b such that $\displaystyle |a|= 6, a^{3}=b^{2}, ba=a^{-1}b$.

What are the generaters a, b for the above group T such that T = <a , b> and what are the elements of T?

2. Originally Posted by aliceinwonderland
There is a nonabelian subgroup T of $\displaystyle S_{3} \times Z_{4}$ of order 12 generated by elements a, b such that $\displaystyle |a|= 6, a^{3}=b^{2}, ba=a^{-1}b$.

What are the generators a, b for the above group T such that T = <a , b> and what are the elements of T?
The order of an element $\displaystyle (x,y)\in S_{3} \times Z_{4}$ is the least common multiple of the order of x in S_3 and the order of y in Z_4.

Elements of S_3 (apart from the identity) have order 2 or 3. Elements of Z_4 have order 2 or 4. So to get an element $\displaystyle a=(x,y)\in S_{3} \times Z_{4}$ to have order 6, you must choose x to have order 3 and y to have order 2.

Next, if a has order 6 then $\displaystyle a^{3}=b^{2}$ must have order 2 and so b must have order 4. So if $\displaystyle b=(z,w)$ then z must have order 1 or 2 in S_3, and w must have order 4 in Z_4.

That should be enough to get you started.