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Let $G$ be a finite group. If for each m$\in$N, $x^m=e$ has at most $m$ solutions in $G$, $G$ is cyclic.

Can you give me a hint of this problem? I don't know.

vonbrand
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user67458
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  • Here are some hints: Take an element of maximal order and consider the subgroup generated by that element. If this is $G$ you are done, so assume not. Show that the subgroup is normal, and that any element not in this subgroup has order coprime to the order of the element you started with. Show that this means that the group is a direct product of cyclic group of coprime orders and thus cyclic. – Tobias Kildetoft Apr 03 '13 at 19:50
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    I'm sure this question has been asked 4857623494523 times already...the last one, btw, just a few days ago. – DonAntonio Apr 03 '13 at 19:52
  • @Tobias Kildetoft // I don't know how to Show that the subgroup is normal, and that any element not in this subgroup has order coprime to the order of the element I started with – user67458 Apr 03 '13 at 20:02

2 Answers2

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Induction on $|G|$. Then every subgroup is cyclic. On another hand, every cyclic subgroup $H$ is normal (else subgroups conjugate with $H$ give us too many solutions of the equation $x^{|H|}=e$). Thus $G$ is either Hamiltonial (then the contrary) or Abelian (then use the decomposition into cyclic subgroups).

Boris Novikov
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    It it were possible to use the decomposition into cyclic groups (I suppose you meant the fundamental theorem of finitely generated abelian groups) then this question is anachronic: anyone knowing the FTFGA would already have seen this simple proof. – DonAntonio Apr 03 '13 at 19:54
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Hints: Let $\,\phi\,$ be Euler's Totient Function, then:

$$(1)\;\;\;\;\;\;\sum_{d\mid n}\phi(d)=n$$

$$(2)\;\;\;\;\text{Define for $\,d\mid n:$}\;\;A_d:=\{g\in G\;;\;ord(g)=d\}\;,\;\;\text{and then show that}\;\;|A_d|=\phi(d)$$

$$(3)\;\;\;\;\text{Supose there exists some $\,d\mid n\,$ s.t.}\;\;A_d=\emptyset . \text{ Get a contradiction to $\,(1)\,$ above}$$

$$(4)\;\;\;\;\;\text{ Prove your claim (where did we use the fact that}\;x^m=e\;\;\text{has at most $\,m\,$ solutions?)}$$

DonAntonio
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