# Show that Let $S= \lbrace f | f:X \to \mathbb{R} \rbrace$ is a group with the multiplication.

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?

• 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? – Mathematician 42 Feb 11 at 9:17

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 $$f$$.