Let $G$ be a group such that for all $a,b \in G$ we have $aba=bab$. Prove that $|G|=1$.

So I have to show that $G =\left\{e \right\} $. Because for any $a,b \in G$ we have $aba=bab$, let $b=e$. Then $aea=eae$ so $a^2 = a$ hence $a = e$. Because $a$ is any we have $G=\left\{e \right\} $.

Does it work?

  • 17
    $\begingroup$ Yes. $\textbf{}$ $\endgroup$
    – Potato
    Aug 16 '13 at 8:12

Your proof works, but it lacks mindless symbol pushing, so I offer two proofs by symbol pushing to restore balance to the world.

Proof I. Write $abab$ in two ways: $babb=aaba$. Then: $$aaba=b(ab)b=abbab$$ And cancel to get $aba=bbab$, but $aba=bab$, so $b=e$. $\square$

Proof II. We start with the natural equation $$baab=aabaa=ababa=babba$$ Now, cancel $ba$ on the left, so that $ab=bba$. But $$aba=bab\implies ab=baba^{-1}$$ Therefore, $bab=bbaa$ and so $ab=baa$ as well. This means $bba=baa$; cancelling on both left and right, $b=a$. Since $a,b$ are abitrary, we are done. $\square$

  • $\begingroup$ Why $b(ab)b=abbab$ ? $\endgroup$
    – Arnaud
    Aug 16 '13 at 9:41
  • 1
    $\begingroup$ @Arnaud I will restate it with more parentheses: $$b(ab)b=(ab)b(ab)$$ $\endgroup$ Aug 16 '13 at 9:46

I make it more detail:

G is a group => e $\in$ G. i.e., |G| >= 1

for some x $\in$ G

1, if $x^{-1}$ not x: |G| >= 3 (at least x, $x^{-1}$, e)

$x^{-1}$x$x^{-1}$ = x$x^{-1}$x, so x=$x^{-1}$

This is to say, for all x $\in$ G, x=$x^{-1}$

2, for any x $\in$ G, since x=$x^{-1}$: x=exe=xex=xe$x^{-1}$=e. so x=e

So G={e}, |G|=1


  • 7
    $\begingroup$ Unfortunately, there is a group in which a nontrivial $x$ is its own inverse. The cyclic group of order $2$ has this property. If you're in a geometry mood, consider the group of symmetries of an isosceles triangle. $\endgroup$ Aug 16 '13 at 8:38
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    $\begingroup$ I'm also puzzled by the claim that nontrivial groups must have order at least $3$. $\endgroup$
    – Potato
    Aug 16 '13 at 8:46
  • $\begingroup$ tt clear that x = e, if x has inverse as x. or what could x be if x = $x^{-1}$ but x != e $\endgroup$
    – gcd0318
    Aug 16 '13 at 10:05
  • 2
    $\begingroup$ Consider the group $\{0,1\}$ with operations $0+0=0$, $1+0=1=0+1$, $1+1=0$. Clearly, this is a non-trivial group of order two. $\endgroup$ Aug 16 '13 at 10:50
  • $\begingroup$ @gcd0318, there are many groups in which there exists at least one element $\;x\ne e\;$ s.t. $\,x^2=e\iff x=x^{-1}\;$ . For example, all the finite groups with even order...Your answer is wrong, so my advice is either correct it or delete it. $\endgroup$
    – DonAntonio
    Aug 16 '13 at 10:56

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