Abstract Algebra – Prove Finite Group of Even Order Contains Element of Order 2

abstract-algebra

Prove that any finite group $G$ of even order contains an element of order $2.$[ Let $t(G)$ be the set $\{g\in G|g\neq g^{-1}\}$. Show that $t(G)$ has an even number of elements and every nonidentity element of $G-t(G)$ has order $2$.]

My proof:

Observe that for any $g\in t(G)$ there exists $g^{-1}(\neq g)\in t(G)$ since every element in a group has an inverse. Thus if there are $n$ elements that belong to $t(G)$ we must have another $n$ elements that belong to $G$. Hence, $t(G)$ has an even number of elements. Now any $g\in G-t(G)$ must also have inverse $g^{-1}(=g)\in G-t(G)$ since all $g\neq g^{-1}$ belong to $t(G)$, so all the elements except the identity in $G-t(G)$ have $g=g^{-1}$ thus $g^2=id$ i.e., every element of $G-t(G)$ has order $2.$ Therefore, we can conclude that there is atleast one element of order $2.$ in $G$.

Can anyone check my proof? Thank you.

Best Answer

The proof has two main problems.

  • When you take an element $g\in t(G)$, you can map it to its inverse, which is again in $t(G)$, as you noted. Since $g\neq g^{-1}$, you see that you have paired the elements of $t(G)$. That allows you to conclude that the order of $t(G)$ is even.
  • You now have to use the fact that both the order of $G$ and of $t(G)$ are even to conclude that there exists an non-identity element not in $t(G)$.
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