# Automorphisms of a particular abelian group

I am trying to solve Exercise 7.9 of An Introduction to K-theory of C*-algebras, It remains for me to show that the Automorphisms of $$\mathbb{Q}\oplus\mathbb{Q}$$ which send $$\{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0)$$ to itself and $$(1,1)$$ to itself are only 2. I was able to find 2 already ,the identity element and the one that sends $$(x,y)$$ to $$(y,x)$$ how do I show that these are the only two?

I don't think what you want is true. Let $$T=\begin{bmatrix} a&1-a\\ c&1-c\end{bmatrix}$$ with $$0 and $$a+a^2\ne c+c^2$$. Then $$T$$ is invertible, $$T\begin{bmatrix} 1\\1\end{bmatrix} =\begin{bmatrix} a+1-a\\ c+1-c\end{bmatrix}=\begin{bmatrix} 1\\1\end{bmatrix},$$ and if $$x,y>0$$ you have $$T\begin{bmatrix} x\\y\end{bmatrix} =\begin{bmatrix} ax+(1-a)y\\ cx+(1-c)y\end{bmatrix},$$ with $$ax+(1-a)y>0$$, $$cx+(1-c)y>0$$.
The examples you found are the cases $$a=1$$, $$c=0$$, and $$a=0$$, $$c=1$$, but there are infinitely many others.
• However, if you insist on the automorphisms to be from $\{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0)$ onto itself, which is probably what is meant in the exercise, then it appears to be indeed true. – Andreas Caranti Mar 19 at 18:26
If $$\alpha = \begin{bmatrix} a&b\\ c&d\end{bmatrix}$$ is such an automorphism, then the fact that fixes $$\begin{bmatrix} 1\\1\end{bmatrix}$$ yields $$a + b = 1 = c + d$$, and applying $$\alpha$$ to $$\begin{bmatrix} 1\\ \epsilon\end{bmatrix}$$, and $$\begin{bmatrix} \epsilon\\1\end{bmatrix}$$, for $$\epsilon \downarrow 0$$, we get $$a, b, c, d \ge 0$$.
Now if $$\alpha$$ maps $$\{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0)$$ onto itself, so does its inverse $$\alpha^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d&-b\\ -c&a\end{bmatrix}.$$ Therefore for $$\epsilon > 0$$ $$\alpha^{-1} \begin{bmatrix} 1\\ \epsilon\end{bmatrix} = \frac{1}{ad - bc} \begin{bmatrix} d - b \epsilon\\-c + a \epsilon\end{bmatrix} \in \{ (x, y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \},$$ so that $$\frac{d - b \epsilon}{ad - bc}, \qquad\frac{-c + a \epsilon}{ad - bc}$$ are both positive. Letting $$\epsilon \to 0$$, we get that $$\tag{pos-neg} \frac{d}{ad - bc}, \qquad\frac{-c}{ad - bc}$$ are both non-negative. But since $$c, d \ge 0$$, one of two numbers in (pos-neg) is $$\le 0$$. Therefore one of $$c, d$$ is $$0$$, and then the other is then $$1$$, as $$c + d = 1$$. The same holds for $$a, b$$ (and clearly if $$c = 0$$, $$d = 1$$, then $$a = 1$$, $$b = 0$$, and conversely), so that we get indeed only the two given automorphisms.