In an abelian group $G,$ let $G^{2} = \{g^2 | g \in G\},$ which is a subgroup. Let $\mathbb{Q}, \mathbb{R}$ and $\mathbb{C}$ denote the rational, real and complex fields, respectively. Let $\mathbb{R}^{+}$ be the positive reals, and $\mathbb{Q}^{+} = \mathbb{Q} \cap \mathbb{R}^{+}.$
$(a)$ Write $\mathbb{C}^{*}$ as a direct product of two subgroups. (Hint: polar form)\
My questions are:
1-I know that I can prove that it is a direct product by either one of those 2 criteria:
{For groups $G,H,K,$ show that the following conditions are equivalent.\
a. $G \cong K \times H.$\
b. $H \triangleleft G, K \triangleleft G, G = HK $ and $H \cap K = \{1\}.$
Am I correct ?
2- If I were to use the first criteria, should I only proof that the function $g : \mathbb{T} \times \mathbb{R}^{+} \rightarrow \mathbb{C}^{*}$ defined by $g(e^{it}, r) = r e^{it}, t\in \mathbb{R}$ is an isomorphism only and this is the complete proof? why we defined $g$ like that?
Let's say $\mathbb{T} = \mathbb{R} / \mathbb{Z} = (0, 1]$. We claim that $\mathbb{T}$ is an additive subgroup of $\mathbb{C}^{*}$. To see this, note if $x, y \in \mathbb{T}$, then $x+y \equiv x+y -\lfloor x+y \rfloor \in \mathbb{T}$, and if $x \in \mathbb{T}$, then $-x \equiv 1-x \in \mathbb{T}$.
To show $\mathbb{R}^{+}$ is a multiplicative subgroup of $\mathbb{C}^{*}$, simply note if $x, y \in \mathbb{R}^{+}$, then $xy \in \mathbb{R}^{+}$, and if $x \in \mathbb{R}^+$, then $\frac{1}{x}\in \mathbb{R}^+$.
Now, the function $g: \mathbb{T} \times \mathbb{R}^{+} \rightarrow \mathbb{C}^{*}$ by $(t, r) \mapsto re^{2\pi i t}$ is easily seen to be a bijective homomorphism, so $\mathbb{T} \times \mathbb{R}^{+} \cong \mathbb{C}^{*}$.