Subgroups of Functions $\mathcal{F}$ is the set of all real-valued functions of a real variable.
I'm trying to show that $H$ is a subgroup of $G$.


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*$G = \langle \mathcal{F}, + \rangle, H = \{ f \in \mathcal{F}(\mathbb{R}): f \text{ is a periodic of period } \pi \}$. 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? 
 A: At the end of (ii), just before the word "Thus", you wrote $\in H$.  You should delete that.  In the first place $-f(x+na)$ is one value if $-f$, whereas it's the function $-f$, not one of its values, that you want to show is in $H$.  Also, before the word "Thus", each equality and each binary relation should be something that you already know is true, whereas $\text{something}\in H$ is what you're trying to prove, not something already known.
Other than that the proof is correct.
I don't like writing "where" when you mean "for every".  The fact is that some somewhat sloppy writers put a trailing "where $m\in\mathcal{Q}$" or something like that when they mean "for every $m\in\mathcal{Q}$", and some write "where $m\in\mathcal{Q}$" when they mean "for some $m\in\mathcal{Q}$", and then the reader is supposed to figure out which, if either, is meant.  Putting "where...." after any sort of mathematical equality or relation or the like should be done when you're trying to explain what your notation or your terms mean, not when you're trying to say "for every" or "for some".
