Let $F$ be the set of all real-valued functions with domain $\Bbb R$ and let $F^{1}$ be the subset of $F$ consisting of those functions that have nonzero value at every point in $\Bbb R$.
Let $G$ be the subset of all $f\in F^1$ such that $f(1)=1$.
Is $\langle G,+\rangle$ a sub-group of $\langle F ,+\rangle$?
Note that $+$ stands for addition.
I could figure out that the identity element of $\langle F,+\rangle $, i.e. $ \mathbf{O}: \Bbb R \rightarrow \Bbb R$ st. $\mathbf O(x)=0 \ \forall x\in \Bbb R$ is not present in $G$. This means that the group $\langle G,+\rangle $ is not a sub-group of $\langle F,+\rangle $.
But my solution says $\langle G,+\rangle$ is not closed in the first place. I couldn't understand how this conclusion could be arrived at.
Can anyone please help?
You just need a counterexample. Take, for example, $f(x) = 1$ and $g(x) = 1 + |x - 1|$. Note that $f(1) = g(1) = 1$ and $g(x) \ge f(x) \ge 1 > 0$, so $f, g \in G$. But, $(f + g)(1) = f(1) + g(1) = 1 + 1 = 2 \neq 1$. Thus $f + g \notin G$. That is, it is possible to add two elements of $G$, but not remain in $G$, meaning $+$ is not a well-defined binary operation when restricted to $G$.
It's not that your argument was wrong, it was more that it was putting the cart before the horse. In order to talk about an identity, we really should make sure we're dealing with a well-defined operation (similarly, before discussing inverses, we must first figure out if there is an identity).