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I want to prove Cauchy's Lemma for abelian groups:

If $G$ is abelian and there exists a prime such that $p$ divides the order of $G$, then there exists a $g \in G$ such that $p=\mathrm{ord}(g)$

I am looking for some tips to prove this (so please no full solutions).

I know that if $G$ is abelian it holds that: $Z(G)=G$ and I know that every subgroup of G is normal in $$G.

Maybe I should consider $U:=\langle g\rangle\subset G$ for some $g \in G$. The order of $U$ should divide the order of $G$, which should be something like $p^rm?$ (Because p divides it)

From now I don't know how to continue (If the way is good so far).

Maybe I should use Sylow's theorem?

Any help is appreciated :)

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Do reduction on order of $G$. As you did, take $U$, when $r>0$, focus on $U$, when $r=0$, consider $G/U$. –  ougao Aug 11 at 13:10

1 Answer 1

Hints:

  1. Prove it for cyclic groups.
  2. Induct on $|G|$ : Take a cyclic subgroup $H$ generated by a non-trivial element of maximal order. If it is all of $G$, apply Part 1. If not, induct. This will take some work because you need to lift an element of order $p$ from a quotient to the big group.
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