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Let $G$ be a finite group. If the order of $G$ is even, prove that there is at least one element $a$ in $G$ such that $a\not= e$ and $a=a^{-1}$.

Here's my idea:

Suppose $\{x_1,\cdots,x_n\}$ is all of the elements in $G$ [edit: I meant to add "such that $x\not =x^{-1}$"] (but without their inverses or the identity $e$). This set has $n$ elements. Now consider the set of the inverses that correspond to each element: $\{x^{-1}_1,\cdots,x^{-1}_n\}$. Since $G$ is a group, there is a one-to-one correspondence between these two subsets. The total order of these groups is $2n$. Consider the identity $e$, which adds 1 to $\left| G \right|$ to give $G=\{e,x_1,\cdots,x_n,x^{-1}_1,\cdots,x^{-1}_n\}$, which yields: $$|G|=2n+1,$$ which is an odd value. But there may be elements that satisfy $y\in G:y=y^{-1}$. In this case, these elements are redundant and so they each add 1 to the order of $G$. Therefore,

$$|G|=[2n_x+1]+n_y$$

The term in brackets is odd, so $|G|$ is even if and only if the number of elements $y$, or $n_y$, is odd. If $n_y$ is odd, it is clear that $\exists y\in G:y\not = e \wedge y= y^{-1}$. Q.E.D.

My questions are:

  1. Is this proof valid?
  2. Any recommendations on how I can make this proof shorter and more elegant? (I struggle with wordiness)

Thank you.

Patrick Shambayati
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  • What do you mean by all of the elements but not their inverses, how do you pick these $x_i$s? – ploosu2 Aug 20 '14 at 18:53
  • Are you starting from the assumption that there is no element which is its own inverse? If not, the definition of ${x_1,x_2,\dots,x_n}$ does not make sense (and I stop reading at this point because I don't expect to understand the remainder of the proof). – JiK Aug 20 '14 at 18:55
  • What I was trying to say was $\forall x \in G \exists x^{-1} \in G$. But now that I'm re-reading it, this doesn't seem very true. Now I'm super lost. :( – Patrick Shambayati Aug 20 '14 at 18:58
  • Ah, yes. When I came up with that first part I was assuming we only consider elements $x:x\not = x^{-1}$. I forgot what was going on in my mind in that first paragraph. – Patrick Shambayati Aug 20 '14 at 18:58
  • See here or here for comments/ideas/whatnot. – Jyrki Lahtonen Aug 20 '14 at 20:12

3 Answers3

4

Your idea is along the right track, but not worded well.

Note that if any element $g$ in $G$ has even order $2k$, then $g^k=g^{-k}$ is your desired element.

Suppose for sake of contradiction that all non-identity elements of $G$ have odd order. Then as you noted, such an element cannot be its own inverse $g=g^{-1}$, else it would have order $2$. Since inverses are unique, you can group all the non-identity elements of $G$ into pairs. Therefore, $|G|$ is odd, a contradiction.

For a more powerful result, see Cauchy's theorem.

angryavian
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2

Assume $x\neq x^{-1}$ for all $x\neq e$.
Using the equivalence relation $\mathcal R$ in $X=G\backslash\{e\}$ defined by $x\mathcal R y$ iff $xy=e$, each class contain two elements, and this relation give a partition of $X$ to a part (of cardinal two), hence the cardinal of $X$ is even so the order of $G$ is odd (contradiction).

Hamou
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I think you've got the essential idea, but might work a bit on the phrasing. The basic thing to notice (as you have) is that the elements $g$ for which $g\neq g^{-1}$ match up in pairs so, removing those, together with the identiy, from the group leaves an odd number of elements, since the order of the group is even. Hence, the remaining set of elements is, in particular, non-empty! But these are the non-trivial elements $g$ for which $g=g^{-1}$.

James
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