$\mathcal{F}$ is the set of all real-valued functions of a real variable.
I'm trying to show that $H$ is a subgroup of $G$.
- $G = \langle \mathcal{F}, + \rangle, H = \{ f \in \mathcal{F}(\mathbb{R}): f \text{ is a periodic of period } \pi \}$. A function $f$ is said to be periodic of period $a$ if there is a number $a$, called the period of $f$, such that $f(x) = f(x + na)$ for every $x \in \mathbb{R}$ and $n \in \mathbb{Z}$.
(i). Let $f, g \in H$, then $f(x) = f(x + n\pi)$ and $g(x) = g(x + n\pi)$ where $n \in \mathbb{Z}$. The sum $[f + g](x) = f(x) + g(x) = f(x + n\pi) + g(x + n\pi) = [f + g](x + n\pi)$. Thus $f + g \in H$.
(ii). Let $f \in H$, then $f(x) = f(x + n\pi)$. The inverse $[-f](x) = -[f(x)] = -f(x + n\pi) = [-f](x + n\pi) \in H$. Thus the inverse $-f \in H$.
So $H$ must be a subgroup of $G$.
Is my proof correct?
Update: How would I go about showing H is not empty? Why is that important? And can I make the assumption that H is not empty?