Tu's An Introduction to Manifolds, in question 15.9 asks a question about the center of $GL(2,\mathbb{R})$. He claims it is isomorphic to $\mathbb{R}^*$.
He then states that the group, $(\mathbb{R}^*, \times)$ has two elements of order $2$, which makes no sense to me as $-1$ is the only one I can find.
Can someone confirm I am correct so I can send an update for the errata?
On
Theorem: No group has exactly two elements of order two.
Proof: Suppose $G$ is a group with exactly two elements of order two. Call them $a$ and $b$. Consider $aba$. We have $$\begin{align} (aba)^2&=(aba)(aba)\\ &=(ab)a^2(ba)\\ &=(ab)(ba)\\ &=ab^2a\\ &=a^2\\ &=e. \end{align}$$
Thus $aba$ has order two (or $aba=e$, but that implies $b=a(aba)a=a^2=e$), so it must equal either $a$ or $b$.
Suppose the former. Then $aba=a$ implies $ab=e$, so $b^{-1}=a$. But the inverse of any order two element is the element itself.
Thus $aba=b$, which implies $ab=ba$. Now
$$\begin{align} (ab)^2&=(ab)(ab)\\ &=(ab)(ba)\\ &=ab^2a\\ &=a^2\\ &=e. \end{align}$$
Hence either $ab=a$ or $ab=b$. But the former implies $b=e$ has order one, whereas the latter implies $a=e$ has order one, which are contradictions.$\square$
You are correct, $\Bbb R^\times$ under multiplication has a single element of order $2$. In fact, if the number of order-$2$ elements in an abelian group is finite, then that number must be $2^n-1$ for some natural number $n$. So it can't be $2$.
Using my powers of mind reading, I induce that the author might have meant that there are two elements that are their own inverse, which is to say, two elements whose square is $1$.