If $G$ is an abelian group, then the inverse of $x$ is equal to $x$, true or false and why?

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    $\begingroup$ Pick your favorite abelian group. Pick your favorite element of it. Perform an empirical check. $\endgroup$
    – Thorgott
    Mar 20, 2020 at 10:21
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    $\begingroup$ It is true for some abelian groups, like the Klein-4 group. But you should understand, jenn, that you should read a question like this as a universal claim: "For all abelian groups G, then for all $x \in G$, $x\in G, $x^{-1} = x$. So my recommendation is that you check more than two of your favorite abelian groups, including cyclic and non-cyclic. $\endgroup$
    – amWhy
    Mar 20, 2020 at 10:35
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    $\begingroup$ Not sufficient, @Thorgott, to pronounce true or false based on one empirical check. If Jenn's favorite group was $\mathbb Z_2$, or the Klein-4 group, she might be lured into believing the statement is true. $\endgroup$
    – amWhy
    Mar 20, 2020 at 10:37
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    $\begingroup$ What exactly did you try? What do you know? Where is this question from? I am voting to close this question because lack of context. $\endgroup$ Mar 20, 2020 at 12:55
  • $\begingroup$ Already a group with $3$ elements refutes this claim. Really not too difficult to find a counterexample here. $\endgroup$
    – Peter
    Mar 21, 2020 at 13:29

1 Answer 1


Obviously false. Consider $\mathbb{Z}/3\mathbb{Z}$ with addition. Then

$$1+2 = 0 = 2+1$$

so $2$ is the inverse of $1$, yet $1 \neq 2$.

And there are plenty more counterexamples.

Note: the converse to your statement does hold.

That is, if every element in a group is equal to its inverse, then the group is abelian. Indeed, if $G$ is such a group and $g,h \in G$, then

$$gh = g^{-1}h^{-1} = (hg)^{-1} = hg$$

An example of a group where every element is equal to its inverse is $\mathbb{Z}_2 \times \mathbb{Z}_2$, as @amWhy notices in the comments.

  • 1
    $\begingroup$ Well said, and I like the observation that the converse of the statement does hold. +1 $\endgroup$
    – amWhy
    Mar 20, 2020 at 10:38

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