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I am working on the problem:

If G is a group of order 2n, show that the number of elements of G of order 2 is odd.

which already has solution (in the link below), please explain the solution in that link

Group of even order contains an element of order 2

Cloud JR K
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    I think it's better to ask clarifications at the linked question instead of starting a new one. – Henno Brandsma Aug 12 '18 at 12:21
  • how to ask such type of question – Cloud JR K Aug 12 '18 at 12:25
  • @JoséCarlosSantos i ask explaination about the answer given in that question... i linked the question (which youcalled duplicate of in my question ... please read it completely. Thanks!} – Cloud JR K Aug 12 '18 at 12:29
  • @CloudJR Nothing that you wrote changes the fact that your question is a duplicate. – José Carlos Santos Aug 12 '18 at 12:30
  • @JoséCarlosSantos but how can i clarify doubts in that answer? after all that question is active approximately 2 years ago. And author of that answer is not actively participated in SE(see his profile) – Cloud JR K Aug 12 '18 at 12:32
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    @CloudJR So, you should reproduce the answer here and then tell us which parts of it you don't understand. – José Carlos Santos Aug 12 '18 at 12:35
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    @ArturoMagidin was last seen 7 hours ago, according to that profile. You can post comments to a posted answer asking for explanations. – Henno Brandsma Aug 12 '18 at 12:36
  • @HennoBrandsma "I remain "gone for the foreseeable future" from math.SE, tex.SE, and meta.SE. Though I may very occasionally post a comment or an answer, these do not signify a desire to re-start my participation.

    Please do not send me private e-mail to call my attention to comments, questions, or other matters related to those sites. Thank you. Also, as I no longer participate in those sites, I do not wish to be sent, by private e-mail, questions that you can just as well ask on those sites. I would have thought it was obvious, but apparently I need to say so explicitly." from his profile

     – Cloud JR K Aug 12 '18 at 12:38
  • @JoséCarlosSantos okay , i will do it next time.. sorry for giving you trouble – Cloud JR K Aug 12 '18 at 12:39

1 Answers1

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OK, the most popular answer defines a relation $x \sim y$ iff $x=y$ or $x=y^{-1}$. You can check that this is indeed an equivalence relation and so we have the set of classes $C_i$, $i \in I$, which form a disjoint partition of $G$. Every class has size $1$ or $2$ (size $1$ iff $C_i = \{x\}$ with $x = x^{-1}$ or equivalently $x^2 = e$. If $x$ is not $e$ and does not have order $2$, it is in a class of size $2$, namely $\{x, x^{-1}\}$. There is at least one class of size $1$, namely $\{e\}$.

If $a$ is the number of classes of size $2$, it has $b:=|I|-a $ classes of size $1$, and $|G| = 2a + b$ and it follows that $b=|G|-2a$ is even, so $b \ge 2$ and so there is at least 1 class besides $\{e\}$ that has size $1$, and this is an element of order $2$.

Henno Brandsma
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