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I am trying to prove something like there exists an element of order $5$ in a group with $625$ elements. How do I do this using basic properties of group theory? Essentially proving cauchy's theorem at the same time.

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    $\begingroup$ But you do know that all orders of elements are divisors of $|G|$? $\endgroup$ – Henno Brandsma Dec 15 '19 at 9:28
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    $\begingroup$ Say you find an element $x$ with order $5^k$ where $k>1$. Then just take $x^{5^{k-1}}$. $\endgroup$ – Berci Dec 15 '19 at 9:35
  • $\begingroup$ Did you only want to consider groups of prime power order, or did you want Cauchy's full theorem on groups of any order? $\endgroup$ – Eric Towers Dec 15 '19 at 9:45
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Let $e \neq g \in G$. Its order is $n$ where $n | 625 = 5^4$ by Lagrange's theorem, so elementary number theory tells us that $n=5^k$ for some $k \in \{1,2,3,4\}$ ($5^0 =1$ is ruled out as $g \neq e$). If $k=1$ we are done. If not, take $g'=g^{5^{k-1}}$ and note that $g'^5 = g^{5^k} = g^n = e$ so we have an element $g'$ of order $5$.

This will work for any group of size that is a positive power of some prime.

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