Answer to Question #144688 in Abstract Algebra for J

Question #144688
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|?
1
Expert's answer
2020-11-17T07:38:09-0500

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."



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