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Given a group $G$ where $g\neq g^{-1}$ for all $g$ other than the identity, show the order of $G$ is odd.

What does it mean for the order of group $G$ to be odd?

Any help welcome.

Ayman Hourieh
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Louise
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  • Imagine you draw the Cayley table, and look at where the group identity entry may lie. Hint: there is one on the diagonal, and the others come in pairs. – Jean-Claude Arbaut Dec 06 '13 at 10:00
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    "The order of $G$ is odd" means that there is an odd number of elements in the group. – Arthur Dec 06 '13 at 10:01
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    Interesting: the tag is "geometry" whereas the question's about groups. These two realms are, of course, very closely connected, yet this is clearer only in rather advanced stuff, which this question definitely belongs not. – DonAntonio Dec 06 '13 at 10:04
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    Not that advanced: geometric groups arise as soon as you are doing affine geometry, and this link with groups happens (at least in France), during the same year that basic properties of groups are taught (roughly, 1st year of BS). – Jean-Claude Arbaut Dec 06 '13 at 10:10
  • See http://math.stackexchange.com/questions/268873/a-finite-group-of-even-order-has-an-odd-number-of-elements-of-order-2 for the contrapositive. – lhf Dec 06 '13 at 10:19
  • Where I studied (Israel) basic group theory is first year/second year. In basic affine geometry the group theory load is rather light, and it really kicks in when Lie groups, geometric group theory (combinatorial group theory) and etc. arrives. Roughly advanced undergraduate-graduate level. – DonAntonio Dec 06 '13 at 14:01

2 Answers2

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Hint (assuming the order is finite):

Pair up each element with its inverse, and remember what's the group's unit's inverse...

DonAntonio
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If $p$ is prime and divides the order of group $G$ then $G$ will contain an element with order $p$ (Cauchy). Note that $2$ is prime. So if the order of $G$ is even then $g^{2}=e$ (or equivalently $g=g^{-1}$) for some $g\in G\backslash\left\{ e\right\} $. This is not the case, so the order of $G$ must be odd.

drhab
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