Number of elements of group with specific order

Question

Consider the group $(G,\cdot)$ where $$G=\left\{\left(\begin{matrix}1&a\\0&b\end{matrix}\right):a,b\in\mathbb{R}, b\neq0\right\}.$$ How many members of $G$ have order 2?

My Attemt

A member $M$ of $G$ will have order two iff $M^2=I$. I.e. $$\left(\begin{matrix}1&a\\0&b\end{matrix}\right)^2=\left(\begin{matrix}1&a+ab\\0&b^2\end{matrix}\right)=I.$$This is true for $b=1$ and $a=0$ in which case $M=I$ or in the case $b=-1$ and $a$ is any real. Hence, there is an infinite number of elements with order 2 in $G$.

Can somebody please verify that this is true?

Your answer is excellent, assuming that you mean "$a,b\in \mathbb{R}$" rather than "$ab\in \mathbb{R}$", as written. Come to think of it, it's right as written also. — vadim123, Nov 19 '13 at 16:19
@vadim123 Yes, thanks, this is what I mean. I corrected it in case anybody wants to have a look at this question in the future. — Zugzwang14, Nov 19 '13 at 16:25

amWhy · Accepted Answer · 2013-11-19T16:30:41.450

2

Very well argued. You are entirely correct.

On the light side: You can stamp your answer Verified @MSE.

edited Nov 19 '13 at 16:30

answered Nov 19 '13 at 16:22

amWhy

209,954

just out of curiosity: what did you edit? I see only 109 characters, but you added 160 characters in body... – draks ... Nov 19 '13 at 20:16

Number of elements of group with specific order

1 Answers1