Show that $\{ f \in G \mid f(2) = 1 \}$ is a subgroup of a group $G$ of functions $\mathbb{R} \to \mathbb{R}^*$ 
Let $G$ be a group of functions of the form $\mathbb{R} \rightarrow \mathbb{R}^{*}$, where the multiplication of $G$ is given by the pointwise multiplication of functions. Let $H = \{ f \in G \mid f(2) = 1 \}$. Prove that $H$ is a subgroup of $G$. 

I have to check that $(fg^{-1})(2) = 1$ for all $f,g \in H$. If this holds then $H$ is a subgroup. But how do I determine $g^{-1}(2)$?
Thanks
 A: For $f,g \in G$, their product (in the group $G = \{\mathbb{R} \to \mathbb{R} - \{0\}\}$) is the function $f \cdot g$ defined by $(f \cdot g) (x) := f(x) g(x)$ for every $x \in \mathbb{R}$. (I'm just writing this to be sure if I understood the definition of the "multiplication" of $G$ correctly)
Let $f \in H$, i.e. $f(2) = 1$. Consider its inverse $\tilde{f}$. It holds that $\tilde{f} \cdot f \equiv I$ (where $I(x) = 1$ for all $x$). Thus for any $x \in \mathbb{R}$ we have $\tilde{f}(x)f(x) = 1$. Inserting $x = 2$, we obtain $\tilde{f} \in H$.
Let now $f,g \in H$, i.e. $f(2) = g(2) =1$. Then we have to evaluate their product $f \cdot g$ at $2$. Thus $(f \cdot g) (2) = f(2) g(2) = 1 \cdot 1 = 1$, i.e. $f \cdot g \in H$.
A: Note $H$ is nonempty since it contains the constant function $f(x)=2$, so that in particular, $f(2)=1$.
Now if $f_1,f_2\in H$, then $$f_1\cdot(f_2)^{-1}=\frac{f_1}{f_2}$$
is a function from $\mathbb{R}$ to $\mathbb{R}^*$ since $f_1$ and $f_2$ are ($f_1$ is never zero, so the fraction isn't either). Lastly: $$f_1(1)\cdot(f_2(1))^{-1}=\frac{f_1(1)}{f_2(1)}=\frac{1}{1}=1$$ so $f_1\cdot(f_2)^{-1}\in H$.
