# Proof of Cauchy Theorem?

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.

• But you do know that all orders of elements are divisors of $|G|$? – Henno Brandsma Dec 15 '19 at 9:28
• Say you find an element $x$ with order $5^k$ where $k>1$. Then just take $x^{5^{k-1}}$. – Berci Dec 15 '19 at 9:35
• Did you only want to consider groups of prime power order, or did you want Cauchy's full theorem on groups of any order? – Eric Towers Dec 15 '19 at 9:45

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 $$5$$.