It is well-known that the unit group of a finite field is a finite cyclic group. But for infinite fields, e.g., $\mathbb{Q}$ or $\mathbb{R}$, the unit groups are not cyclic. I heard this fact in my class and I'm trying to prove it.
Assuming the unit group is infinite cyclic with the generator $a$, I tried to show some element, $\frac{1}{a+1}$ or $\frac{1}{a^2+1}$ cannot be a power of $a$, but it doesn't seem hopeful. Could someone help me?
Note that in fact, more is true: a finite subgroup of the multiplicative group of any field is cyclic.
If the field $K$ has characteristic zero, it contains $\mathbf{Q}$ as a subfield, and thus $K^*$ contains $\mathbf{Q}^*$ as a subgroup. Hence it is sufficient to show that $\mathbf{Q}^*$ is not cyclic (because a subgroup of a cyclic group is cyclic). Suppose $a$ is a generator of $\mathbf{Q}^*$. Then also $a/2 \in \mathbf{Q}^*$, but no power of $a$ can equal $a/2$ (consider absolute values).
If $K$ has characteristic $p$ and is an algebraic extension of $\mathbf{F}_p$, for every element $a \in K$ you can consider the finite field $\mathbf{F}_p(a)$ to see that the subgroup generated by $a$ must be finite.
Finally if $K$ is a transcendental extention of $\mathbf{F}_p$, is contains $\mathbf{F}_p(t)$ as a subfield, so $K^*$ contains $\mathbf{F}_p(t)^*$ as a subgroup, and again it suffices to show that $\mathbf{F}_p(t)^*$ is not cyclic. Again, for any purported generator $a$, consider $a/2$, but this time you argue on its degree.
