$\mathbb C$ isomorphism to $\mathbb{R} \times \mathbb{R}$ under multiplication How can I show, that $(\mathbb C,\cdot)$ is not isomorphic to  $(\mathbb{R},\cdot) \times  (\mathbb{R},\cdot)$ under multiplication?
I tried to point out that $f(1) = 1$, then pair $(1,1) \rightarrow 1 + 0i$, but then I god stuck. 
I know these semigroups are isomorph under addition, but any reasonable mapping seems not to be even homomorphism (due to fact, that $(a + bi)(c + di) \neq a+c + i(b+d)$.
Edit: Sorry, I assumed those were groups.
 A: In the semigroup $(\mathbb{C}, \cdot)$ each element admits a square-root, that is for each $a \in \mathbb{C}$ there is an $x$ such that $x^2 = a$. 
Suppose there is an isomorphism $f$, that is a bijective map such that $f(cd)=f(c)f(d)$ for all $c,d$. Let $a = f^{-1} ((-1,-1))$ and let $x$ be a squaroot of $a$. Let $f(x)= (b_1,b_2) \in \mathbb{R}^2$. Then, on the one hand $f(a) = (-1,-1)$, yet on the other hand $f(a) = f(x^2) = f(x)^2 = (b_1,b_2)^2 = (b_1^2, b_2^2)$. 
This would yield $b_1^2 = -1 $ and $b_2^2 = -1 $ in the real numbers, which is not possible. This contradiction shows that there is no isomorphism.
A: To show that $(\mathbb{C}, \cdot)$ and $(\mathbb{R}, \cdot)^2$ are not isomorphic as semigroups, observe that both semigroups have a (necessarily unique) annihilator, an element $a$ such that $ax = a$ for any $x$. Both semigroups have a unique element $e$ such that $ex = x$ for any $x \not= a$ (in fact the elements other than $a$ form a subsemigroup that is actually a group). The equation $x^2 = e$ has 2 solutions in $(\mathbb{C}, \cdot)$ but 4 solutions in $(\mathbb{R}, \cdot)^2$. Since we can define $a$ and $e$ using only the semigroup operation, the two semigroups cannot be isomorphic. 
