In any finite group, number of elements not equal to their own inverses is even number
In my book they have paired elements with their inverses, being elements and inverses different from each other. How do i see this hint ? Thanks
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1You seem to have answered your own question: you can divide such elements into pairs: $(a,a^{-1})$. – almagest Jun 15 '16 at 05:50
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@almagest You can also try and prove it by contradiction-if I got puzzled on an exam and blanked out,that's what I'd try. – Mathemagician1234 Jun 15 '16 at 05:52
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@almagest so i have to count both $a$ and $a^{-1}$ – Gathdi Jun 15 '16 at 05:54
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@almagest if there are $k$ pairs then $2k$ elements are there – Gathdi Jun 15 '16 at 05:57
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@Gathdi If $a\ne a^{-1}$ then $a$ and $a^{-1}$ are different elements, so yes you have to count them both! – almagest Jun 15 '16 at 05:58
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@almagest but this is not satisfying. How can i be sure that group is partitioned into k pairs – Gathdi Jun 15 '16 at 06:01
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It isn't, there might be any number of elements which are their own inverse. But the elements which are not their own inverse are grouped into pairs. – almagest Jun 15 '16 at 06:03
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@almagest Thanks – Gathdi Jun 15 '16 at 06:09
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@almagest why cannot be it that there are 3 elements {a,b,c} such that inverse of a is b and inverse of b is c – Gathdi Jun 15 '16 at 06:14
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@Gathdi Maybe because inverses are unique, if $b$ is the inverse of $a$, then $a$ is the inverse of $b$. – sqtrat Jun 15 '16 at 06:17
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@sqtrat hmm. this is quite confusing – Gathdi Jun 15 '16 at 06:19
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1$b=a^{-1},b^{-1}=c$. Multiplying the lhs together and the rhs together we get $1=a^{-1}c$. Multiplying by $a$ we get $a=c$. – almagest Jun 15 '16 at 06:20