# Subgroup of $\left(GL_2\left(R\right),\:\cdot \right)$

Is $t\left(GL_2\left(\mathbb{R}\right)\right)=\left\{x\in GL_2\left(\mathbb{R}\right)|\:ord\left(x\right)\:<\:\infty \right\}$ a subgroup of $\left(GL_2\left(\mathbb{R}\right),\:\cdot \right)$ ? Justify your answer.

$GL_2\left(\mathbb{R}\right)$ is the general linear group, meaning it's the group of 2 x 2 invertible matrices of real numbers.

How do I solve this problem? I'm really new to group theory.

• @quid Ah yes, of course, I am sorry. This link is better, I hope. The answer of "amWhy" includes the one of Omnomnomnom. – Dietrich Burde Jun 9 '16 at 17:35

The answer here is going to be no. For example, consider the product $$\pmatrix{1&0\\0&-1} \pmatrix{1&1\\0&-1} = \pmatrix{1&1\\0&1}$$ can you see how this "counterexample" shows that the set fails to be a subgroup?
• So I basically needed to find two elements from $GL_2\left(\mathbb{R}\right)$ which have an order that is less than infinite, but for which their product has an infinite order? Hence their product wont be part of $GL_2\left(\mathbb{R}\right)$, right? – MikhaelM Jun 9 '16 at 17:23
• It's not that their product won't be a part of $GL_2(\Bbb R)$ (of course they will be a part of $GL_2(\Bbb R)$ since $GL_2$ is a group). It's that their product won't be a part of $t(GL_2)$ (which is probably what you meant anyway). – Omnomnomnom Jun 9 '16 at 17:27
• So yes: we had to find two matrices of finite order whose product has infinite order. In so doing, we show that $t(GL_2)$ fails to be a subgroup. – Omnomnomnom Jun 9 '16 at 17:28