Groups with Complex Numbers I'm working on a quick proof related to the complex numbers.
Take $G$ to be the group of all nonzero complex numbers under
multiplication ($\circ$). Let $H$ be the set of of complex
numbers such that the sum of squares of the two real parts of the 
complex number is equal to 1. I want to show that $H$ under
multiplication is a subgroup of $G$. I can see how to do the
closure part of the proof, but I am having issues finding an
accurate inverse for each element, likely due to the fact that
I am uncertain about the identity element in the complex case.
 A: The identity under multiplication is 1.  The recriprical (multiplicative inverse) of $x + yi$ is $\frac {x}{x^2 + y^2} - \frac {y}{x^2 + y^2}i$.  
A: Remember that the identity under multiplication is 1. So the inverse of $z \in \mathbb C$ is the reciprocal $\frac{1}{z}$. Since $|z| = 1$, we know $\left|\frac{1}{z}\right| = \frac{1}{1} = 1$.
We can compute this explicitly if you are not comfortable with the absolute value of a complex number. Let $z = x+iy$ such that $x^2+y^2=1$ and $x, y \in \mathbb R$. We now have $$\frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2}.$$
Of course, we need to check that sum of squares of real and complex parts is 1. So we get
$$\left(\frac{x}{x^2+y^2}\right)^2 + \left(\frac{-y}{x^2+y^2}\right)^2= \frac{x^2+y^2}{(x^2+y^2)^2},$$ which is 1 because we assumed that $x^2+y^2=1$.
A: Here is a helpful remark for the future: a group and any of its subgroups must have the same identity element. If it were otherwise, this would contradict the uniqueness of the identity element of the parent group.
Since it's not clear which identity you're unsure about (that of $G$ or $H$) we can look at why it must be $1$ for both.
Because $(\Bbb R \setminus \{0\}, \times)$ is a subgroup $G = (\Bbb C \setminus \{0\}, \times)$ and $1$ is the multiplicative identity of $\Bbb R$ (I hope this is something you're very comfortable with!), you can rest assured that it is also the identity element of your group $G$.
Then, knowing that $1$ is the identity of $G$, if $H$ is going to be a subgroup of $G$ it had better have $1$ as its identity element as well. Just verify that this is so if you're not sure, but it should be pretty ingrained that $1z = z$ for all $z \in \Bbb C$.
