In a finite group $G$ if every element is of some prime power order (prime may vary with element) and if $G$ has non trivial center then prove that $G$ is actually of prime power order. Deduce that any group G of order pq (where p and q are distinct primes) with non trivial center is cyclic.
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2 Answers
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Hint. If an element $x$ of order $p$ exists in $Z(G)$, it commutes with any other element $y$ of order $q$ in $G$. What's that say about the order of $xy$?
answered Apr 12, 2013 at 17:51
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$\begingroup$ @GArunkumar does that contradict the hypothesis? what if $p=q$? $\endgroup$– Alexander Gruber ♦Apr 13, 2013 at 2:02
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$\begingroup$ ya. so there cannot be an element of order q. so only conclusion we can get is G is p-group. am I correct? so such a need not be cyclic is your point. is it? $\endgroup$– GA316Apr 13, 2013 at 3:16
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$\begingroup$ @GArunkumar Yes, you are correct, if every element is $p$-power then $G$ is a $p$-group, and that is all you know. (It doesn't have to be true that $G$ is abelian, though.) $\endgroup$– Alexander Gruber ♦Apr 13, 2013 at 5:50
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Let $g\neq 1$ be in the center, and have prime order divisible by $p$.
Suppose there is another element $h$ with prime power divisible by $q\neq p$. Since $h$ and $g$ commute, the order of $hg$ has to be the least common multiple of the orders of $h$ and $g$, but that is clearly not prime power. Hence no such $h$ exists in $G$, and $G$ is a $p$ group. As such it has order a power of $p$.
If $G$ had order $pq$ but was not cyclic, then the only possible orders of elements are $p$ and $q$ and 1 (but of course, the identity is the only thing with order 1): these are all prime powers (with $1=p^0$, I suppose). If the center were nontrivial, then by the first part $pq$ is a power of a prime: this is a contradiction. Hence $G$ must be cyclic.
answered Apr 12, 2013 at 17:40