Question

Assume that $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$ and ab=ba. If $\mathbf{\left|}\mathbf{a}\mathbf{\right|}$ and $\mathbf{\left|}\mathbf{b}\mathbf{\right|}$ are relatively prime, prove that ab has order $\mathbf{\left|}\mathbf{a}\mathbf{\right|}\mathbf{\left|}\mathbf{b}\mathbf{\right|}$ .

Short answer

It is proved that, ab has order$\left|a\right|\left|b\right|$.

Step-by-step solution

Referring to the result of Exercise 31

${\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}}^{\mathbf{\left|}\mathbf{a}\mathbf{\right|}\mathbf{\left|}\mathbf{b}\mathbf{\right|}}\mathbf{=}\mathit{e}$

Proving that ab has order |a| |b|

As we know, from the property of group elements ${\left(ab\right)}^{\left|ab\right|}=e$

And rewrite it as:

${\left(ab\right)}^{\left|ab\right|}=a^{\left|ab\right|}{b}^{\left|ab\right|}=e$

For any arbitrary $i,j\in N$, we have $a^i{b}^j=e$.

This implies $a^i=b^{-j}=e$ and we know that $b^j$ and $b^{-j}$have the same order.

Since role="math" localid="1654263758629" $a^i$has a divisor of $\left|a\right|$, $b^j$ has a divisor of $\left|b\right|$ and $\left(\left|a\right|\left|b\right|\right)=1$, we can conclude that $\left|a^i\right|=\left|b^j\right|=1$.

Therefore, role="math" localid="1654263941006" $\left|a\right|$and role="math" localid="1654263944487" $\left|b\right|$ both divides role="math" localid="1654263947863" $\left|ab\right|$.

As it is given that role="math" localid="1654263982065" $\left|a\right|$and role="math" localid="1654264011140" $\left|b\right|$are relatively prime.

Hence, we can conclude that ab has order $\left|a\right|\left|b\right|$.