**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.4**

Prove that the multiplicative groups $\mathbb{R}^\times$ and $\mathbb{C}^\times$ are not isomorphic.

Solution: Recall from Exercise 1.6.2 that isomorphic groups necessarily have the same number of elements of order $n$ for all finite $n$.

Now let $x \in \mathbb{R}^\times$. If $x = 1$ then $|x| = 1$, and if $x = -1$ then $|x| = 2$. If (with bars denoting absolute value) $|x| < 1$, then we have $$1 > |x| > |x^2| > \cdots,$$ and in particular, $1 > |x^n|$ for all $n$. So $x$ has infinite order in $\mathbb{R}^\times$.

Similarly, if $|x| > 1$ (absolute value) then $x$ has infinite order in $\mathbb{R}^\times$. So $\mathbb{R}^\times$ has 1 element of order 1, 1 element of order 2, and all other elements have infinite order.

In $\mathbb{C}^\times$, on the other hand, $i$ has order 4. Thus $\mathbb{R}^\times$ and $\mathbb{C}^\times$ are not isomorphic.