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In any finite group $G$, the number of elements not equal to their own inverse is an even number.

I'm not sure how to go about this. I was considering assuming the number of elements not equal to their own inverse is an odd number (contradiction), but I'm not sure how to get a contradiction. I was considering thinking about when $|G|$ has even order and odd order, and then finding a way to do it directly, but I'm not sure. I think that in a finite group the number of elements equal to their own inverse should be even since they come in pairs. Any hints or solutions are greatly appreciated.

ddswsd
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HINT: $a\ne a^{-1}$ if and only if $a^{-1}\ne(a^{-1})^{-1}$

Brian M. Scott
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  • Is it because $a \ne a^{-1}$ come in pairs, implying it must be even? If the number was odd, then we would come across an $a$ that would have to be equal to it's inverse? – ddswsd Jan 04 '21 at 19:45
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    @ddswsd: Yes, it’s because they come in pairs. – Brian M. Scott Jan 04 '21 at 19:46
  • Wow, that's so obvious. I feel silly lol thanks for helping. – ddswsd Jan 04 '21 at 19:47
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    @ddswsd: You’re very welcome. (It happens to all of us from time to time!) – Brian M. Scott Jan 04 '21 at 19:48
  • Please strive not to add more dupe answers to FAQs. – Bill Dubuque Jan 04 '21 at 20:20
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    @BillDubuque: I might very well have offered this hint even if I’d known that this was a FAQ and been able to find the earlier answers quickly, since they are complete. Besides, when a very quick response is possible, I’d rather give it than waste time searching. Time enough to search later and link to earlier answers; a variety of answers is often a good thing anyway. – Brian M. Scott Jan 04 '21 at 20:30
  • If you truly believe the answer offers something novel compared to the many prior answers then please add it to one of the earlier threads (vs. here). – Bill Dubuque Jan 04 '21 at 20:36