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$\begingroup$

a,b are elements of the group G

I have no idea how to even start - I was thinking of defining a,b as two square matrices and using the non-commutative property of matrix multiplication but I'm not sure if that's the way to go...

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    What are $a$ and $b$?2012-11-10
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    Perhaps the tag group-theory implies these matrices need to be invertible.2012-11-10
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    Sorry a,b are elements of a group G2012-11-10
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    I don't know if I'm even supposed to use matrices - that's just the first thing I thought of.2012-11-10
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    Manasa, the answers below were based on assuming $a$ and $b$ are just matrices before your comment added context. The answer is still no, and you could still show it using 2-by-2 matrices, but invertible ones instead: try it! (Try to choose $a$ and $b$ so that their squares commute with everything, even though they don't commute with each other.)2012-11-10
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    It certainly implies $a^2b^2=b^2a^2$.2012-11-10
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    There is also a counterexample in a group with $6$ elements.2012-11-10
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    @JonasMeyer, do the matrices have to be invertible due to the inverse property of groups?2012-11-10
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    @Manasa: If you're looking for a counterexample with matrices, then they do have to be invertible. In the future, please specify what $a,b$ actually are.2012-11-10
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    Why do they have to be invertible?2012-11-10
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    @Manasa: I'm assuming that you want the group operation to be matrix multiplication. In that case, every element needs a multiplicative inverse because that's part of the definition of a group.2012-11-10
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    Matrix multiplication just happened to be the first thing I thought of but I don't know if that's actually how I'm supposed to approach it. And thanks for clarifying, I thought that might be the reason.2012-11-10

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