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Let G be a group with even order. (no binary operation given)

Prove using only the definition of a group that there is $ a\in G$ s.t $a \neq e $ where $a= a^{-1}$

Originally I wanted to say if G is even then order of G =2k for some integer k and somehow show that some element with order k and order 2 must exist. sadly the group isnt nicely defined for me to do this.

Ive been trying to find a contradiction since G is a group if $a \in G$ then $a^{-1} \in G$ since G is finitely ordered the maximum order an element can have is the order of G and $ \forall a $ where $a\in G$ $a^2 \neq e$

EDIT: I'm not sure who thinks this is the same as the linked question but it using many items what i have not been taught nor does it even really answer the stated question the comment below the answer does however.

Faust
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    Hint: Suppose not. Then for every element $g\neq e$ form the pair ${g,g^{-1}}$. Count the order of $G$ in terms of the number of pairs. – lulu Jan 24 '17 at 20:36
  • omg thats so simple, thanks! – Faust Jan 24 '17 at 20:37
  • About your edit: The linked question asks for a slightly stronger claim that trivially implies the given one. And the answer is basically identical to the one in the comment once you see what the terms mean (these are terms one need to get familiar with to study group theory anyway). – Tobias Kildetoft Jan 24 '17 at 20:44

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Cauchy's Theorem says that a finite group $G$ of even order has an element of order $2$, with $p\mid |G|$ and $p=2$. Then $a^2=e$, but $a\neq e$, i.e., $a=a^{-1}$.

Dietrich Burde
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