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. $\blacksquare$

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

1 Answer 1


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


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