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If $G$ is a group such that $|G|=2n$. Prove that there's an odd number of elements of order 2, and then there's an element which is its own inverse, besides of the identity.

If we consider all the elements of $G$ that have order different than 2, we have two cases: 1) the elements with order greater than 2, and 2) elements with order 1.

For 1). Let $A=\{g\in G: g^k=e, k>2\}$, then if $g\in G$ then $g^{-1}\in G$ (should I prove that the inverse has the same order?), and also $e\not\in G$. Then there's an even number of elements in A, because $g\not = g^{-1}$. So $|A|=2m$ for some $m$.

For 2). The only element of order 1 is the identity, so $B=\{g\in G: g^k=e, k=1\}=\{e\}$ then $|B|=1$.

So $|A|+|B|=2m+1$ like we wanted, but if that happens, the second part of the problem doesn't make sense.

So doing this in a different way, instead of directly counting the elements of B, we consider the complement of A, so $|A^c|=2k$, that way $|G|=|A|+|A^c|=2n$ checks out right. But $A^c=\{g\in G: g^k=e, k\not>2\}=\{g\in G: g^k=e, k=1 or 2\}$, so because the identity is the only element of order 1, then there must be at least one element in $A^c$ of order 2, that means $g_o^2=g_o*g_o=e$ hence $g_o=g_o^{-1}$.

Is this right? What am I doing wrong in the first try?

I know this problem has been solved before in this site, but I really wanted to make it on my own, sorry if this is redundant.

Asinomás
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Abodi
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  • possible duplicate of Group of even order contains an element of order 2 – vadim123 Jan 25 '14 at 01:05
  • @vadim123 I know, please don't close my question, I wish people could help me with what I wrote, plus the link you shared uses a different approach, and I just want to know what I did wrong and why, thanks. – Abodi Jan 25 '14 at 01:12
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    Your proof is correct for part one. Just to clarify some of the reasoning, if g does not have order two, then g^{-1} != g (since then g^2 = 1), so these are in fact two distinct elements. You should mention this when counting elements, otherwise you might be counting some elements more than once. – breeden Jan 25 '14 at 01:17
  • @breeden there might be a way to know if I'm counting more? – Abodi Jan 25 '14 at 01:21
  • @Abodi, I guess we could be a little more careful with counting. You have $A = { g \in G | ord(g) > 2 }$. This is a finite set. Take $g \in A$, and note that $g^{-1}$ is in $A$ (easy). Note that $g \neq g^{-1}$ so we have two elements in $A$. Remove these two elements from $A$. If $A$ is now empty, you are done and have removed an even number of elements. Otherwise, take another element from $g \in A$. Note that $g^{-1}$ has not already been removed (by uniquness of inverses) and so $g^{-1}$ is now in $A$. Take these two elements out.. and so on – breeden Jan 25 '14 at 01:26
  • Voted to close as a duplicate over the earlier objection because the OP is satisfied and the question is (ultimately) a duplicate. Thus the question (while initially valuable) no longer serves a purpose. – dfeuer Jan 25 '14 at 03:56

1 Answers1

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Your first try is correct. By your proof, you know the number of elements of order 2 is odd, hence can not be zero. This means there exists at least one, say $g$, which is just the element you wanted in the second part.

Wei Zhou
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