SOLUTION TO PROBLEM H-14

It is enough to show that a and b commute. Rewrite aba=b³ as

b^-1ab=b²a^-1

So
$$\begin{multline*}b^{-2}ab^2=b^{-1}(b^{-1}ab)b= b^{-1}(b^2a^{-1})b=b^2(b^{-1}a^{-1}b)=\\ b^2(bab^{-1})^{-1}=b^2(b^2a^{-1})^{-1}=b^2ab^{-2} \end{multline*}$$
In other words

ab⁴=b⁴a

Since b⁵=1 or b⁴=b^-1, this means that ab^-1=b^-1a. Therefore

ba=ab

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Apr 26 07:49:32 CDT 2002