Is $H = \{ z \in G \mid o(z) < \infty \}$ a subgroup of $G=(\Bbb C^*,\cdot)$? 
Consider $G = ( \Bbb C^* , \cdot)$ and  $H = \{ z \in G \mid o(z) \; \text{is finite} \}$ . Will $H$ be a subgroup of $G$?

The lecture I'm referring to says that this will be a subgroup of $G$. All elements having modulus $1$ are of finite order, however, if I take  $(1/\sqrt{2}+(1/\sqrt{2})i)$ and $(1/\sqrt{2} -(1/\sqrt{2})i)$ , they both belong to H but their sum doesn't.
So is my conclusion accurate that $H$ will not be a subgroup of $G$?
 A: You are right the sum is not in $H$. But the group operation in question is multiplication of complex numbers, not addition.
It is true that $H = \{z \in \mathbb{C} \setminus \{0\} : z \text{ has finite  order} \}$ is a subgroup of $(\mathbb{C} \setminus \{0\}, \cdot)$. Checking this is a straightforward exercise.
A: Use the one-step subgroup test.
Since $o(i)=4$, we have $i\in H$. Hence $H\neq\varnothing$.
By definition, $H\subseteq G=(\Bbb C^*,\cdot)$.
Let $x,y\in H$. Then $m=o(x), n=o(y)$ are both finite. Now
$$\begin{align}
(xy^{-1})^{mn}&= x^{mn}(y^{-1})^{mn}\\
&=(x^m)^n(y^{-n})^m\\
&=1^n1^m\\
&=1,
\end{align}$$
so that $o( xy^{-1})$ divides $mn$. But $mn$ is finite. Hence $o(xy^{-1})$ is finite. Hence $xy^{-1}\in H$.
Hence $H\le G$.
A: It's clear that $1 \in H$, and by the definition of $H$ we have $ H \subset \mathbb{C}^*$. 
For all $x$ in $H$ we denote the order of $x$ by $o(x)$.
Let $x, y \in H$, we have
$$ (xy)^{o(x) o(y)}=x^{o(x) o(y)}y^{o(x) o(y)}\\ =(x^{o(x)})^{ o(y)}(y^{o(y)})^{o(x)}\\ =1^{ o(y)}1^{o(x)}\\ =1$$
then $xy \in H$. 
Let $x \in H$ and let $x^{-1}$ be the inverse element of $x$ in $ (\mathbb{C}^*,.)$, we have
$$1=1^{o(x)}=(xx^{-1})^{o(x)}=x^{o(x)}(x^{-1})^{o(x)}=1(x^{-1})^{o(x)}=(x^{-1})^{o(x)}$$.
then $x^{-1} \in H$. 
Hence, $H$ is a subgroup of $ (\mathbb{C}^*,.)$.
