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Assume that we have a group together with inverse elements for all its group members, the closure property and also the identity. Can we follow from these properties that our group also has to fulfill associativity?

EDIT: Or to reformulate my question: If I can set up an arbitrary multiplication table which is complete and closed, so that in each of its rows and columns every element of the "group" is appearing once and only once, does this define me a group automatically? This is really puzzling me..

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    $\begingroup$ No. There are counterexamples. What you're referring to is called a loop. The octonians under multiplication are an example of a nonassociative loop. $\endgroup$
    – Jared
    Apr 29, 2013 at 18:24
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    $\begingroup$ I'm pretty sure the answer is "no". A definition this fundamental and long-lived would not have extraneous components. Have you tried to prove that a set with a closed binary operator with inverses and identities must be associative? I don't think is actually answers your question, but it is interesting and related: math.stackexchange.com/questions/172694/… $\endgroup$ Apr 29, 2013 at 18:24
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    $\begingroup$ @Jared: But as far as I see, the set of unit octonions for example is not closed: en.wikipedia.org/wiki/Octonions#Definition $\endgroup$ Apr 29, 2013 at 19:19
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    $\begingroup$ @ToddWilcox I have tried to set up some counterexamples as you suggested, but I always end up having an associative set :-) $\endgroup$ Apr 29, 2013 at 19:23
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    $\begingroup$ We need not take the unit octonians. Take all of them under multiplication. Or, if you'd like, take the unit octonians along with their additive inverses for a loop of order $16$. $\endgroup$
    – Jared
    Apr 29, 2013 at 21:16

2 Answers 2

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No. Let $A=\left\{0,1,2\right\}$ with the operation $+:A\times A\rightarrow A$ given by $0+0=0$, $0+1=1$, $0+2=2$, $1+0=1$, $1+1=0$, $1+2=1$, $2+0=2$, $2+1=2$, $2+2=0$. Then $0$ is the (bilateral) identity, the (bilateral) inverse of $x$ is $x$, but $$(1+2)+1=1+1=0\quad\text{and}\quad 1+(2+1)=1+2=1$$ hence $(A,+)$ is not associative.

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  • $\begingroup$ Thank you very much! Can such a counterexample also be formulated for a multiplicative group law? $\endgroup$ Apr 29, 2013 at 18:44
  • $\begingroup$ The $+$ sign in this example is just a symbol. If you like, replace it by $\cdot$. There is no mathematical distinction between an "additive" group law and a "multiplicative" group law, except that one normally only uses the plus sign for an abelian group. $\endgroup$
    – Lee Mosher
    May 4, 2013 at 14:58
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There are lots of loops which are not groups, for example the multiplicative structure of a non-associative algebra such as the octonions $\mathbb{O}$ or any interesting Lie algebra.

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  • $\begingroup$ I feel the answer is tangential to the topic, at least somewhat. $\endgroup$
    – jiten
    Jan 14, 2018 at 5:41

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