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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.

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Since $g=g^{-1}$ for all $g\in t(G)$, then you cannot say "Observe that for any $g\in t(G)$ there exists $^{−1}(\neq g)\in t(G)$ [...]" – M Turgeon Aug 29 '12 at 19:22
Do you want your definition to be $$t(G) = \lbrace g \in G \vert g \neq g^{-1}\rbrace ?$$ – Kris Williams Aug 29 '12 at 19:23
@DrKW You're probably write, since then any non-dentity element not in $t(G)$ has order two, as stated. – M Turgeon Aug 29 '12 at 19:24
A less inspired argument: if the order of your group is even, there is a non-trivial $2$-Sylow subgroup, so you might just as well suppose that the group is a $2$-group. Now such a thing has a non-trivial center, so we may just as well assume that our $2$-group is abelian. And now it is easy :-) – Mariano Suárez-Alvarez Aug 29 '12 at 19:32
@Legendre: well, using Cauchy's theorem to prove a special case of Cauchy's theorem is a bit circular :-) – Mariano Suárez-Alvarez Aug 29 '12 at 20:13
up vote 2 down vote accepted

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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There appears to be a mistake in the (hint-part?) question itself: $t(G) = \{g\in G|g = g^{-1} \text{, i.e. } g^{2} = 1\}$ is exactly the set of elements of $G$ which have order 1 or 2, and so every non-identity element of $t(G)$ has order 2, and no element of $G-t(G)$ has order 2.

The hint should read: [Let $t(G)$ be the set $\{g\in G|g=g^{-1}\}$. Show that $G-t(G)$ (and thus also $t(G)$) has an even number of elements and every nonidentity element of $t(G)$ has order 2.]

For your proof, the first line contradicts the definition of $t(G)$; you can't have $g\in t(G)$ with $g^{-1}\neq g$. Try switching the roles of $t(G)$ and $G-t(G)$, and it seems to me that your proof is mostly correct.

In your last step, you say that every element of $t(G)$ has order 2 when I think you meant to say every non-identity element has order 2. Given this result, though, why can we conclude that there is an element of order 2 in $G$? In other words, what is it that guarantees $t(G)$ has more than just the identity element? You still need to use the fact that $|G|$ itself is even!

As a side-note, while it isn't necessary to go into this much detail for a rigorous proof, try to show (for yourself) why it must be that if there are $n$ distinct elements in $G-t(G)$ that their inverses actually provide another $n$ distinct elements. That is, if $a,b\in G-t(G)$ and $a\neq b$, why must $a^{-1}\neq b^{-1}$?

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Sorry for the typo. – Lyapunov Aug 29 '12 at 22:22

If we use the definition $$t(G) = \lbrace g \in G \vert g \neq g^{-1} \rbrace,$$ then please use my suggestions.

I have an issue with the statement "Thus if there are $n$ elements that belong to $t(G)$ we must have another $n$ elements that belong to $G$." How can you conclude from there being $n$ elements in $t(G)$ anything about $G$?

To show that $t(G)$ is even, you may want to find a way to pair all elements of $t(G)$ and show that each pair contains distinct elements of $t(G)$.

Finally, where do you show that $G - t(G)$ contains anything except the identity?

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@Jason: It does once the typo in the problem statement is corrected. – Brian M. Scott Aug 29 '12 at 19:33
@Brian: Fair enough. – Jason DeVito Aug 29 '12 at 19:36

The following argument is not different in essence from many of those above, but may be more succinct: At least one element of $G$ is its own inverse (the identity). If no other element of $G$ is its own inverse, then the non-identity elements of $G$ may be partitioned into mutually inverse pairs, so $G$ has odd order.

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