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Let $S= \lbrace f | f:X \to \mathbb{R} \rbrace$, where $X$ is a non empty set, and $\mathbb{R}=\lbrace x\in \mathbb{R} | x>0 \rbrace$.

For every $(f,g) \in S \times S$, let $f.g$ be the function $f.g:X \to \mathbb{R}^+$ defined as $f.g (x)=f(x).g(x), \forall x \in X$. Show that $(S,.)$ is a group.

My attempt:

Let $f,g,h \in S$, we have that

1) $f.g:X\to \mathbb{R}^+$ So the operation “.” Is binary operation on $S$.

2) From the definition of the operation “.” , we have that $f.g:X \to \mathbb{R}^+$, and since $f.g \in S$ [from (1)] then $(f.g).h:X \to \mathbb{R}^+$. Similarly, $f.(g.h):X \to \mathbb{R}^+$. Hence the domain of $f.(g.h)$ is equal to the domain of $(f.g).h$. Now, we have that

$((f.g).h) (x)=(f.g)(x). h(x)$

$=(f(x).g(x)).h(x)$

$=f(x).(g(x).h(x))$

$=f(x). (g.h)(x)$

$=(f.(g.h))(x)$.

Thus, the operation “.” is associative on $S$.

3) The identity element, let $I:X \to \mathbb{R}^+$, such that $I(x)=1$ for all $x\in X$. We have that $f.I:X \to \mathbb{R}^+$

$(f.I)(x)=f(x).I(x)$

$=f(x).1$

$=f(x)$

Similarly, $I.f:X \to \mathbb{R}^+$

$(I.f)(x)=I(x).f(x)$

$=1.f(x)$

$=f(x)$

4) What about for the inverse?

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    $\begingroup$ If $X$ is a singleton, the set of functions from $X$ to $\mathbb{R}^+$ can be identified with $\mathbb{R}^+$, the inverse of an element $x$ is simply $\frac{1}{x}$ (which exists as $x\neq 0$). Can you extend this to functions? $\endgroup$ – Mathematician 42 Feb 11 at 9:17
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In view of the inverse, let $f:X\rightarrow{\Bbb R}_{>0}$. Then $g:X\rightarrow{\Bbb R}_{>0}$ with $g(x) = 1/f(x)$ is the inverse of $f$.

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