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Is there some group with elements $a,b$ such that $\langle ab\rangle=\langle ba\rangle$, yet the cyclic subgroups generated by $a,b$ and $ab$ are distinct?

If $\langle ab\rangle=\langle ba\rangle$ then $ab$ and $ba$ do commute, that is $ab^2a=ba^2b$, but I cannot find a useful consequence of this equality.

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  • $\begingroup$ If $a$ and $b$ are two distinct 2-cycles in $S_3$, then both $ab$ and $ba$ are 3-cycles generating $A_3$. As pointed out by others the Klein 4-group give an even smaller example. $\endgroup$ – Jyrki Lahtonen May 23 '14 at 6:37
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How about the Klein group $K=<a, b\;|\;a^2=b^2=a^{-1}b^{-1}ab=1>$?

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  • $\begingroup$ Since the condition makesit necessary that $1,a,b,ab$ are distinct, this is clearly the smallest counterexample ... $\endgroup$ – Hagen von Eitzen May 23 '14 at 5:50
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Take the quaternion group of order 8: $Q=\{1,-1,i,-i,j,-j,k,-k\}$, and put $a=i$ and $b=j$.

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$(\mathbb{Z}_2\times \mathbb{Z}_2,+)$ should do the trick.

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  • $\begingroup$ See jpvee's answer $\endgroup$ – Hagen von Eitzen May 23 '14 at 6:03
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Somewhat at the other extreme to jpvee's minimal example is the "largest minimal" example

$$G_{m,n}=\langle \,a,b\mid (ab)^n=ba, (ba)^m=ab\,\rangle $$ for suitable $n,m\in\mathbb Z\setminus \{0\}$. However, it is not always immediately clear that elements of a presented group that look distinct really are distinct. But we can easily identify one of these: $$ G_{1,1} =\langle \,a,b\mid ab=ba\,\rangle\cong \mathbb Z\oplus \mathbb Z.$$

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