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Here's what I'm trying to prove.

Let $(G,\circ)$ be a finite group. Define a set $S \subset G$:

$$S = \{x \in G: x \neq x^{-1}\}$$

where $x^{-1}$ is the inverse of $x$ in $G$ with respect to the given operation. We can divide $S$ into pairs that contain an element and its inverse. Prove that $|S|$ is even.

Proof Attempt:

Suppose that $|S|$ is odd. Then, there exists a pair in $S$ that contains only one element. We call it $a$. Since every pair is supposed to contain an element and its inverse, it follows that $a \circ a = e$, where $e \in G$ is the identity element. However, that is equivalent to asserting that $a = a^{-1}$. This is a contradiction. Hence, $|S|$ is even.

Does the proof above work? If it doesn't, why? How can I fix it?

Mousedorff
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  • Rather than using a pair with repeated elements, a better way to state it is if there are an odd number of elements, then there exists two distinct elements that share the same inverse, which is not possible from the definition of an inverse – Dhanvi Sreenivasan May 27 '20 at 04:37
  • Oh, so the wording I've used is a bit off but the idea is fine? – Mousedorff May 27 '20 at 04:38
  • Yeah, I mean your argument isn't wrong in any way. It's just how i would write it – Dhanvi Sreenivasan May 27 '20 at 04:40
  • Ahhh I see, I see. Alright, that makes sense. Thank you so much. – Mousedorff May 27 '20 at 04:40
  • The statement generalizes easily to showing the any involution on a finite set will have an even number of non-fixed points. Inversion on any group is an example of an involution, and its fixed points (excluding the identity), of course, are also called "involutions". – Geoffrey Trang May 27 '20 at 05:09
  • I'll take note of that and try to prove that on my own. I'll also have to do some research on those terms because Pinter hasn't really introduced them yet. Actually, the above result that I've proved is from an exercise. – Mousedorff May 27 '20 at 05:11

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So for each such $x$ we can pair it with its inverse, $x^{-1}$. If we take a different $x$, we get a different $x^{-1}$. Thus the number of them is even.