I would like to check if following sets are groups with multiplying in $\mathbb{C}$. I show my attempt, however I am not sure about my solutions. Could you check it ?
a. $\mathbb{R}$
b. $\{z\in \mathbb{C} : |z|= 1\}$
c. $\{n\in \mathbb{N} : n > 0\}$
neutral element is $1$.
a. No, because $0$ has no an inverse element.
b. Yes, $0\notin \{\cos(\alpha)+i\sin(\alpha) : \alpha\in[0,2\pi]\}= \{z\in \mathbb{C} : |z|= 1\}$. Each element has an inverse element, set is closed under multiplication (it is easy to check that if ($|z_1|=1$ and $|z_2|=1$ then $|z_1z_2|=1$).
c. Yes, because $0\notin \{n\in \mathbb{N} : n > 0\}$, closed under multiplication.$
Am I ok ?
a) You are correct, and your argument is correct. You can easily see that there is no $x\in\mathbb R$ such that $0\cdot x=1$.
b) You are correct, but you still have to show that the set is closed under inversion. So, if I take an element from the set, is the inverse element also in the set?
c) Think again. Just because the set is closed under multiplication, that doesnt' make it a set. What is the inverse of $2$? Is it a member of the group?