# Non-torsion elements

• Jun 8th 2009, 07:03 AM
Swlabr
Non-torsion elements
Do these form a subgroup? That is to say, as the inverse of a non-torsion element (an element of non-finite order) is non-torsion, basically the question is: If we multiply two non-torsion elements together do we get another non-torsion element?
• Jun 8th 2009, 02:51 PM
siclar
Is the identity a nontorsion element?
• Jun 8th 2009, 03:28 PM
NonCommAlg
siclar has already answered the question. a less trivial question is to find examples of two torsion elements $\displaystyle a,b$ of a group such that $\displaystyle ab$ becomes non-torsion.
• Jun 8th 2009, 06:20 PM
Jose27
Quote:

Originally Posted by NonCommAlg
siclar has already answered the question. a less trivial question is to find examples of two torsion elements $\displaystyle a,b$ of a group such that $\displaystyle ab$ becomes non-torsion.

$\displaystyle A= \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$ and $\displaystyle B= \left( \begin{array}{cc} 0 & 1 \\ -1 & -1 \end{array} \right)$ then $\displaystyle A^4 = I = B ^3$ but $\displaystyle AB$ has infinite order.
• Jun 8th 2009, 10:37 PM
Swlabr
Quote:

Originally Posted by siclar
Is the identity a nontorsion element?

Sorry - do the non-torsion elements along with the identity form a subgroup?
• Jun 8th 2009, 11:44 PM
NonCommAlg
Quote:

Originally Posted by Swlabr
Sorry - do the non-torsion elements along with the identity form a subgroup?

still no! for example in the group of $\displaystyle 2 \times 2$ invertible matrices over $\displaystyle \mathbb{Q}$ let $\displaystyle a=\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\displaystyle b=\begin{pmatrix}1 & 0 \\ -2 & 1 \end{pmatrix}.$ then both $\displaystyle a,b$ are non-torsion but $\displaystyle ab=\begin{pmatrix}-1 & 1 \\ -2 & 1 \end{pmatrix}$ is torsion because $\displaystyle (ab)^4=I.$