1

Let $H$ be a finite group of order $n$ with the property that for every $d\mid n, x^d = 1$ has at most $d$ solutions. Then $H$ is cyclic.

Let $\alpha = \{h'_1, h'_2, ..., h'_k\}$ be the set of all orders of elements. If $n \in \alpha$ done. Assume $n \notin \alpha.$ Consider $\mathrm{lcm}(h'_1, ..., h'_k) = h'.$ We know that there exists an element $g$ such that $o(g) = h'.$ Since $n \notin \alpha, h' \neq n.$ Hence, we have that for all $x \in H, x^{h'} = 1.$ However, we have $n$ solutions where $h' < n.$ Contradiction. It must be that $n \in \alpha.$

Does this work?

user26857
  • 52,094
green frog
  • 3,404
  • 1
    How do you know there is an element of order $h'$? – lulu Apr 28 '17 at 21:20
  • 1
    For another approach look at the responses to this question – lulu Apr 28 '17 at 21:26
  • Here is a hint: for every such $d$, count the number of elements with this order. – David Wheeler Apr 29 '17 at 23:00
  • Oh for every such $d$ the number of elements is either 0 or $\phi(d).$ And since we know that $\sum_{d | n} \phi(d) = n,$ none of those values can be 0 otherwise contradicting that the group is of order $n.$ Hence, we have that the number of elements of order $n$ is nonzero so cyclic. – green frog Apr 29 '17 at 23:28

0 Answers0