# All elements of a set have a multiplicative inverse

I was reading my groups notes, and was wondering if this is true --

Claim: If all elements in a set $S$ have a multiplicative inverse then the set is closed under multiplication.

Proof: Let $x,y\in S$. Then $xy \in S$ because $(xy)^{-1}$ is in the set and therefore $xy=((xy)^{-1})^{-1}$. Thus $S$ is closed under multiplication. $\blacksquare$

• How can you say that $(xy)^{-1}$ belongs to $S$ if you don't know yet that $xy$ does? Jan 22, 2016 at 16:53
• Right, so this is a circular argument. Thanks, seemed to good to be true :) Jan 22, 2016 at 16:55
• @Pierre-GuyPlamondon actually, wouldn't $(xy)^{-1}$ be in the set because $(xy)^{-1} = y^{-1}x^{-1}$ and $x, y \in S \implies x^{-1}, y^{-1} \in S$? Jan 23, 2016 at 1:05
• You cannot say that $x^{-1}, y^{-1} \in S$ implies $y^{-1}x^{-1}\in S$ if you don't already know that $S$ is closed under multiplication. Jan 23, 2016 at 9:05

Just consider the subset of $(\mathbb{R^*},\times)$ $$S=\{\frac{1}{2},2\}$$ and note that $1\not\in S.$