0
$\begingroup$

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.

  • 1
    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. – 2017-01-24
  • 0
    omg thats so simple, thanks! – 2017-01-24
  • 0
    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). – 2017-01-24

1 Answers 1

1

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., $a=a^{-1}$.

  • 0
    Cauchy seems way overkill here given the context. – 2017-01-24
  • 1
    Yes, this is true. But it cannot be too early to learn about this Theorem in the beginning of group theory. The same holds for Lagrange Theorem, which is related. – 2017-01-24
  • 0
    Sure, but working out exercises like this one using more elementary methods is also a good way to get familiar with how groups "work". – 2017-01-24