Prove that the set of complex numbers of the form e^iθ where 0<θ<2pi forms a group under multiplication?

I think you must have the inequality wrong. At the moment, that set isn't even closed under multiplication.

However, if we consider the set

G = {e^(iθ) | 0 <= θ <= 2pi }

then it does form a group with multiplication.

Take two elements in G, e^(ia) and e^(ib), where 0 <= a,b <= 2pi.

e^(ia) * e^(ib) = e^(i(a+b))

Now we have two cases: a + b <= 2pi or a + b > 2pi.

In the first case, we're fine, since 0 <= a + b <= 2pi.

In the second case, we need to show that 0 <= a + b - 2pi <= 2pi.

Then we have e^(i(a+b)) = e^(i(a+b-2pi)) * e^(i(2pi)) = e^(i(a+b-2pi)).

This shows that G is closed under multiplication.

From here, it's fairly direct.

Multiplication of complex numbers is associative, so multiplication in G must be associative.

e^(i(0)) * e^(ix) = e^(i(0+x)) = e^(ix), so e^(i(0)) is the identity element.

Finally, e^(ix) * e^(i(2pi - x)) = e^(i(x+2pi-x)) = e^(i(2pi)) = 1 = e^(i(0)).

so every element in G has an inverse in G.

It is not a group because in the interval with its endpoints excluded

0< θ <2π

the unit element is missing :

1 = exp(i0)