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Show that in any group of order 100 , either every element has order that is a power of a prime or there is an element of order 10. Without using Sylow's theorem how to approach this problem ? By Lagrange's theorem , possible orders of elements of G are 1,2,4,5,10,20,25,50 and 100. Any hint or suggestion will be appreciated.

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    $\begingroup$ What if there's an element of order $20$? $\endgroup$ – Lord Shark the Unknown Jun 17 '18 at 11:15
  • $\begingroup$ Thanks i got it $\endgroup$ – blue boy Jun 17 '18 at 12:30
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Hint: You're on the right track. The list contains only $10, 20, 50, 100$ which are not prime powers. If e.g. $x$ has order $50$, then $x^5$ has order $10$.

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  • $\begingroup$ Thanks . I got it. $\endgroup$ – blue boy Jun 17 '18 at 12:29
  • $\begingroup$ If |x|= 20, x^ 2= 1 $\endgroup$ – blue boy Jun 17 '18 at 12:29
  • $\begingroup$ Likewise it can be solved. $\endgroup$ – blue boy Jun 17 '18 at 12:30

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