Alg7.11

From Exampleproblems

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 $m_1=m\,$ .