# If $G$ is an abelian group, then inverse of $x$ is equal to $x$? [closed]

True/False

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

• Pick your favorite abelian group. Pick your favorite element of it. Perform an empirical check. Mar 20, 2020 at 10:21
• 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$. So my recommendation is that you check more than two of your favorite abelian groups, including cyclic and non-cyclic. Mar 20, 2020 at 10:35 • Not sufficient, @Thorgott, to pronounce true or false based on one empirical check. If Jenn's favorite group was$\mathbb Z_2$, or the Klein-4 group, she might be lured into believing the statement is true. Mar 20, 2020 at 10:37 • 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. Mar 20, 2020 at 12:55 • Already a group with$3\$ elements refutes this claim. Really not too difficult to find a counterexample here. Mar 21, 2020 at 13:29

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.

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