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$\mathcal{F}$ is the set of all real-valued functions of a real variable.

I'm trying to show that $H$ is a subgroup of $G$.

  1. $G = \langle \mathcal{F}, + \rangle, H = \{ f \in \mathcal{F}(\mathbb{R}): f \text{ is a periodic of period } \pi \}$. 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?

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I would say that in part (ii) you need to change $a$ fo $\pi$, and you are missing the final step of going from $-f(x+na)$ to $[-f](x+na)$, thus showing $[-f]$ is periodic of period $\pi$. Also, you forgot to verify that $H$ is not empty. – Arturo Magidin Aug 9 '11 at 20:10
Nearly. In (ii) you’ve left off a step: $-f(x+na)=[-f](x+na)$ is needed to finish showing that $-f$ is periodic, and $a$ should be $\pi$. Also, (i) and (ii) could be vacuously true, so you need to check that $H\ne\varnothing$. By the way, it’s sufficient to show that $f-g \in H$ whenever $f,g \in H$; that’s equivalent to the conjunction of your (i) and (ii). – Brian M. Scott Aug 9 '11 at 20:13
@Gary: In (i) he simply has to show that for any fixed $n$, $[f+g](x+n\pi)=[f+g](x)$, so it is the same $n$. ‘Inverse’ here refers to the inverse in the group $G$; this is an additive group, and the inverse is simply the negative of the original function, which certainly is defined on all of $\mathbb{R}$ – Brian M. Scott Aug 9 '11 at 20:20
@gary: Your last comment is irrelevant. If $f$ and $g$ are in $H$, then $f(x) = f(x+n\pi)$ for all integers $n$, and $g(x)=g(x+n\pi)$ for all integers $n$. In order to show that $f+g$ is in $H$, you don't need to show that $\pi$ is the smallest period, you just need to show that $f+g$ is periodic with period $\pi$; that is, you need to show that for any (particular) integer $n$, $(f+g)(x+n\pi) = (f+g)(x)$. – Arturo Magidin Aug 9 '11 at 20:37
@Jon: Yes, that is exactly what I mean: just say "$H$ is not empty, because the function give explicit example is in $H$". That proves $H$ is not empty. Yes, $K$ is empty because no real valued function can satisfy the given condition. But that means that the implication $(f,g\in K)\Rightarrow (f+g\in K)$ is true (because the antecedent is always false), and likewise $(f\in K)\Rightarrow (-f\in K)$ is true (because the antecedent is always false). If you read the definition of "subgroup" carefully, as well as the associated theorems, you will see non-emptiness is always there. – Arturo Magidin Aug 9 '11 at 20:49
up vote 3 down vote accepted

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".

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Your second paragraph isn’t quite right: Arturo and I pointed out a couple more small flaws that you missed. – Brian M. Scott Aug 9 '11 at 20:26

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