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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.

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

As for "how to solve it": my best advice is that if there isn't an obvious way to prove it, start by looking for a counterexample. Either you find one, or you start to see why there aren't any.

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  • $\begingroup$ 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? $\endgroup$ – MikhaelM Jun 9 '16 at 17:23
  • $\begingroup$ 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). $\endgroup$ – Omnomnomnom Jun 9 '16 at 17:27
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Jun 9 '16 at 17:28
  • $\begingroup$ Yup, that's what I meant. Thank you for your time :) $\endgroup$ – MikhaelM Jun 9 '16 at 17:31

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