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This question references an intermediary step in the proof of:

Prove that any finite group $G$ of even order has an element of order $2$,

which the proof can be found here.

Specifically the opening lines of the proof say "Pair up if possible each element of $G$ with its inverse..." I'm not entirely sure what it could mean if it were not possible to do so. If $G$ is a group, then surely both $g, g^{-1} \in G$. Otherwise $G$ wouldn't be a group by definition.

Can anyone provide any guidance as to why we need to specify if possible?

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Decaf-Math
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    If $g$ is its own inverse then ${g,g^{-1}}$ is not a pair. – lulu Jan 24 '17 at 01:29
  • @lulu, is it to say we reach into the "bag" $G$, and since there are only a finite number "things" in $G$ that we can physically pair it with that we call it a pair (since we already grabbed one item, that item is no longer in the bag meaning we can't pair it with itself)? – Decaf-Math Jan 24 '17 at 01:31
  • Since the problem concerns parity, it makes a big difference if the nominal pair contains $2$ or only $1$ element. The logic of the proof in this case is that $e\in G$ is clearly it's own inverse...so if every other element was part of a unique pair then the group would have odd order. – lulu Jan 24 '17 at 01:33
  • I don't understand your comment about a finite number of things. We know the element $g$ has an inverse, as $G$ is a group. The only question is whether or not $g^{-1}$ is distinct from $g$. if it is, then we can make a pair. Otherwise we can't. – lulu Jan 24 '17 at 01:35
  • Okay. I think I see what the issue is here. We say that ${1,2,3,3,4,5}$ is not a set since the element $3$ repeats (in fact, it is a multiset in that case). Hence ${g,g^{-1}}$ would not be a pair because if $g = g^{-1}$, then we'd have ${g,g^{-1}} = {g,g}$ which is a multiset.

    I had to consult the definition of a "pair" in terms of math.

     – Decaf-Math Jan 24 '17 at 01:36
  • Right. We're trying to count the elements of the group. It's no good if you count some elements multiple times. The idea is that $#G$ is twice the number of pairs plus the number of self-inverse elements. Hence there must be evenly many self-inverse elements. – lulu Jan 24 '17 at 01:38

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To summarize and actually resolve this question:

If $G\setminus\{e\}$ has no elements of order $2$, then each element in the set can be paired uniquely to its inverse, so $|G\setminus\{e\}|$ is even, and $|G|$ is odd.

The contrapositive of this proposition is the one you were trying to prove.

rschwieb
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Because the point of the proof is the goal to show that a group with even order has an element of order 2, which means that $g = g^{-1}$, therefore it cannot be uniquely paired up with its own inverse (because you'd end up pairing it with itself which violates the way we pair up elements with their inverses).

Since $|G| = 2n$, then you are left with $\{e, g\}$ where $g = g^{-1}$, so $g^2 = e$ necessarily. An excellent proof by existence without construction.

q.Then
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