Is a set $A=\left\{ z \in C: \left| z\right| = 2\right\}$ an abelian group?

Question

There is a set $A=\left\{ z \in C: \left| z\right| = 2\right\}$, and operation $\cdot$ which for every pair of complex numbers $(z,w)$ assigns result $z \cdot w = \frac{zwi}{2}$ , where i is imaginary unit.

1) Check if $\left( A, \cdot \right)$ is an abelian group
2) Check if $\left( C, \cdot \right)$ is a group

I am not sure how to do this but I checked the conditions for a group for 2) because I don't know how to limit my analisis to set $A$ - for example how to tell if $\left( A, \cdot \right)$ is internal operation and I found that for x=0 inverse element doesn't exist.

Answer 1

Since you found that for $x=0$ the inverse does not exists, you have answered 2) : $(\mathbb{C} , \ \cdot \ )$ is not a group...

For 1) : first indeed, you need to check if $\cdot$ is internal : if you take two elements of $A$, does the product belongs to $A$?

For the unit element : a unit element of $A$, since the group is abelian, is an element $e$ (in fact unique) such that : $$\forall z\in A, \qquad z\cdot e=z \quad \Leftrightarrow\quad ze\frac{i}{2}=z \quad \Leftrightarrow\quad e=-2i \tag{since $0\notin A$}$$

For the inverse element let $z\in A$ ; we are seeking $h\in A$ such that $z\cdot h = -2i$, that is $\frac{zhi}{2}=-2i \Rightarrow h=\frac{-4}{z}$. So, $$ \forall z\in A, \qquad z^{-1}:=\frac{-4}{z}$$

Answer 2

$x=0$ is not an element of $A$! in fact, $A$ is a group. You need to show the following facts

1. If $z,w\in A$ then $z\cdot w \in A$ (the operation is internal)
2. There exists $e\in A$ such that for all $z\in A$ $z\cdot e = z \in A$. (there exists a unit element)
3. for all $z\in A$ there exists $z^{-1}\in A$ such that $z\cdot z^{-1}=e$.(there exists an inverse)
4. for all $z,w\in A$ $z\cdot w = w\cdot z$. (the group is abelian)

Solution/hints

1. $|z\cdot w| = |\frac{zwi}{2}|=1/2|z||w||i|=1/2\cdot 2 \cdot 2\cdot 1 = 2$ and so $z\cdot w\in A$
2. take $e=-2i$
3. take $z^{-1} = -\overline z$
4. calculation.

As you said $(\mathbb{C},\cdot)$ is not a group because $0$ has no inverse.

Answer 3

It would be better if you write $z = 2e^{i\theta}$, then for $z_1 = 2e^{i\theta_1}$ and $z_2 = 2e^{i\theta_2}$ we have that:

$$z_1 \cdot z_2 = \frac{4e^{i\theta_1}e^{i\theta_2}e^{i\frac{\pi}{2}}}{2} = 2e^{i\left(\theta_1 + \theta_2 + \frac{\pi}{2}\right)}$$

Now it's not hard to prove that $2e^{-i\frac{\pi}{2}} = -2i$ is the identity element, the inverse of $z = 2e^{i\theta}$ is $2e^{-i\left(\theta + \frac{\pi}{2}\right)}$. Also the operation is commutative and associative as real numbers are under addition. Hence $A$ is an abelian group. And finally trivially $|z_1 \cdot z_2| = 2$, hence $(A,\cdot)$ is an abelian group.

In the second case your deduction is alright. If $(\mathbb{C}, \cdot)$ is group, then $(A,\cdot)$ is a subgroup and we must have that $-2i$ is the identity in $(\mathbb{C}, \cdot)$, too. But as you've noted, then $0$ doesn't have an inverse. In fact the only bad point is $0$, as $(\mathbb{C} - \{0\}, \cdot)$ is a subgroup and this has to do something with the fact that $0$ can't be represented in polar form, i.e. the trick we used for part $a)$.