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Let $a,\,b,\,c$ be elements of some group with $abc=e$. Does it follow that $bca=e$? And does it follow that $bac=e$? Give proof or counterexample.

If $abc=e$ then $bc= a^{-1}$ then it follows that $bca= (a^{-1})a=e$. But how about $bac=e$? Can someone help me. I'm self learning the fundamentals of abstract algebra.

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Your proof that $bca = e$ is correct.

Let us see what happens if $bac$ is the identity. Now, $b(ca) = e \implies ca = b^{-1} $ and $b(ac) = e \implies ac = b^{-1}$, hence $ac = ca$, i.e. they commute.

This helps us find a counterexample. We need to find some elements so that they multiply to give the identity, but don't commute with each other. One example are permutations. Take, for example, $a = (123), b= (23), c = (12)$

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