Alg7.11

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Let a\, and b\, belong to the group G\,. If ab=ba\, and |a|=m,|b|=n\,, where m\, and n\, are relatively prime, show that |ab|=mn\, and that <a>\cap <b>={1}\,.

Proof:

(ab)^{{mn}}=(a^{m})^{n}(b^{n})^{m}=1\implies |ab|{\bigg |}mn\,

\implies |ab|=m_{1}n_{1}\, where m_{1}|m,n_{1}|n\,

So a^{{m_{1}n_{1}}}b^{{m_{1}n_{1}}}=1\,.

If m=m_{1}m_{2}\, then raise both sides of the last equation to the power m_{2}\, to get b^{{mn_{1}}}=1\implies |b|=n{\big |}mn_{1}\,. Since (m,n)=1\,, n{\big |}n_{1}\,. But already n_{1}{\big |}n\, so n_{1}=n\,.

If n=n_{1}n_{2}\, then raise both sides of the last equation to the power n_{2}\, to get a^{{m_{1}n}}=1\implies |a|=m{\big |}m_{1}n\,. Since (m,n)=1\,, m{\big |}m_{1}\,. But already m_{1}{\big |}m\, so m_{1}=m\,.

Therefore |ab|=m_{1}n_{1}=mn\,.

c\in <a>\cap <b>\implies c^{{|a|}}=c^{m}=1,c^{n}=1\,

\implies |c|{\big |}m,|c|{\big |}n{\mathrm  {\,and\,}}(m,n)=1\implies |c|=1\implies c=1_{G}\,.


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