Answer to Question #144688 in Abstract Algebra for J

Question

Assume that G is a finite group, and that there exist g,h are element of G, g doesnt equal h, such that order(g)=p, order(h)=q, where p and q are distinct primes. What can be concluded about |G|?

Answer 1

It can be concluded that $|G|\ge p\cdot q.$ Indeed, $G$ contains the subset $H=\{g^nh^m\ |\ 0\le n<p, 0\le m<q\}$. If $g^nh^m=g^kh^s$ for some $0\le n,k<p$ and $0\le m,s<q$, then $g^{n-k}=h^{s-m}$ where $-p< n-k<p$ and $-q< s-m<q$. Since $order(g)=p$ and $order(h)=q$ , where $p$ and $q$ are distinct primes, each non-identity element of the cyclic subgroup $\langle g\rangle$ has order $p$ and each non-identity element of the cyclic subgroup $\langle h\rangle$ has order $q$. Thus, we conclude that $g^{n-k}=e$ and $h^{s-m} =e$, and consequently the inequalities $-p< n-k<p$ and $-q< s-m<q$ imply $n-k=0$ and $s-m=0$. Therefore, $n=k$ and $s=m$, and we conclude that $|H|= p\cdot q.$