In this exercise, you will show using contradiction that $R^∗$ is not cyclic. Suppose that it is cyclic and let $g ∈ R^∗$ be a generator. Then $R ^∗ = <g>$. In particular, $|g|^ {1/2} ∈ R ^∗ $ and so $|g| ^{1/2} = g ^m $ for some integer m. Show that the only solutions to this equation are $g = ±1$. Where’s the contradiction?
I dont understand why $|g| ^{1/2} = g^m$ implies $g = ±1$. I also dont know where the contradiction is.
(Assuming $R^*$ is the group of real numbers (except $0$) with multiplication).
We are assuming that $g$ generates $R^*$. We know there's a nonzero real number $h$ satisfying $h = \sqrt{|g|}$. If $R^*$ is cyclic then $g^m = h$ for some integer $m$.
That implies that $g^{2m} = h^2 = |g|$. Taking absolute values, we have$|g|^{2m} = |g^{2m}| = \big||g|\big| = |g|$, and $|g|^{2m-1} = 1$. The only real roots of $1$ are $\pm 1$, so $|g| = 1$. Thus $|g|^n = 1$ for all integer $n$.
If $R^*$ is cyclic, then we just proved that every element of $R^*$ has absolute value equal to $1$, which is absurd.