# Alg7.11

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

Proof:

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

$\displaystyle \implies |ab|=m_1 n_1\,$ where $\displaystyle m_1|m, n_1|n\,$

So $\displaystyle a^{m_1 n_1} b^{m_1 n_1} = 1\,$ .

If $\displaystyle m=m_1 m_2\,$ then raise both sides of the last equation to the power $\displaystyle m_2\,$ to get $\displaystyle b^{mn_1}=1\implies |b|=n\big|mn_1\,$ . Since $\displaystyle (m,n)=1\,$ , $\displaystyle n\big|n_1\,$ . But already $\displaystyle n_1\big|n\,$ so $\displaystyle n_1=n\,$ .

If $\displaystyle n=n_1 n_2\,$ then raise both sides of the last equation to the power $\displaystyle n_2\,$ to get $\displaystyle a^{m_1 n}=1\implies |a|=m\big|m_1 n\,$ . Since $\displaystyle (m,n)=1\,$ , $\displaystyle m\big|m_1\,$ . But already $\displaystyle m_1\big|m\,$ so $\displaystyle m_1=m\,$ .

Therefore $\displaystyle |ab|=m_1n_1=mn\,$ .

$\displaystyle c\isin \cap\implies c^{|a|}=c^m=1, c^n=1\,$

$\displaystyle \implies |c|\big|m, |c|\big|n \mathrm{\,and\,} (m,n)=1 \implies |c|=1 \implies c=1_G\,$ .