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Suppose $G$ is a finite Abelian group and, $\forall\ n\in \mathbb{N} $, there exist at most $n$ elements in $G$ which satisfy $x^n=1$. Prove $G$ is cyclic.

Thanks for your help.

Bhaskar Vashishth
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Ivan S. Guerra
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  • I'm not sure.. can I take $n$ the number of elements and then take an element of $G$ that satisfies the equation and use it as generator of the cyclic group?? – Ivan S. Guerra Aug 28 '15 at 22:36
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    That statement doesn't make sense. Do you mean less than or equal to? – pancini Aug 28 '15 at 22:39
  • Hint: What must finite Abelian groups look like? – Christopher A. Wong Aug 28 '15 at 22:40
  • I think you mean "less than or equal to n elements", no? Assuming that, then the argument is essentially the same as in the standard proof that any finite subgroup of a field is cyclic, which you can find, eg, here: http://math.stackexchange.com/questions/59903/finite-subgroups-of-the-multiplicative-group-of-a-field-are-cyclic – lulu Aug 28 '15 at 22:45

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Lets prove it by contradiction.

Let $G$ is not cyclic and finite abelian, we can find an element $g\in G$ which has maximal order among elements of $G$, say order $k$ and $k < |G|$ . Now pick an element $h\in G$ \ $<g>$ and let $|h|=m$ then obviously $m \le k$ as $k$ is maximal among orders of elements of $|G|$.

Now consider $l=\text{lcm} (k,m) $. As $G$ is abelian there exist an element of order $l$ and obviously $l \ge k$, so it implies by maximality of $k$, that $l=k$. But $l=k \implies m|k\ $ that is $k=tm$ for some positive integer $t$. Now note that all elements $\{1,g,g^2,\dots g^{k-1}\} \cup \{h\}$ satisfies $x^k=1$, thus more than $n$ elements satisfies $x^k=1$.

Hence Proved

Bhaskar Vashishth
  • 11,279