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What's the correct converse of this problem?

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Which is the correct converse for the pictured question (so you can see the exact wording)?

__Converse One__: Take the converse of the statement *agb=cgd* always implies that *ab=cd*. That is:

**Suppose ***G* is a group with the property that *ab=cd* always implies *agb=cgd*. Prove or disprove *G* is abelian.

This seems like a kind of silly statement, and I see no way to prove it. The only way to make *ab=cd* is if *a=c* and *b=d* or if *a=b=c=d*.

__Converse Two__: Take the converse of the question. That is:

Suppose that G is an abelian group. Prove that *G* has the property that *agb=cgd* always implies *ab=cd*.

Which is too easy to be part (b) of this question. If G is abelian and *agb=cgd*, then we have

*agb=cgd*

*gab=gcd*

*g*^{-1}gab=g^{-1}gcd

ab=cd

Which means that *G* has that property.

TIA.

Re: What's the correct converse of this problem?

the wording of the question is vague, and i understand your confusion. however, i favor your "converse two" as the correct one.