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Given the following Cayley Table (where e is the identity element):

Cayley Table in question

How would I go about proving that the table does not form a group?

I have checked closure, identity, inverses, and all 27 combinations of associativity excluding the ones that include the identity element.

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    $\begingroup$ It's surprisingly difficult to prove things that are false! $\endgroup$ – Derek Holt Feb 8 '15 at 22:05
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    $\begingroup$ "The only group in Z4 is the Klein 4-group." This statement makes no sense, and I don't know what you mean. For example, you might mean "the only group (up to isomorphism) with 4 elements is the Klein 4-group." That statement is definitely false, but at least it is understandable. Or if this is not what you mean, perhaps you could elaborate? $\endgroup$ – mathmandan Feb 8 '15 at 22:11
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    $\begingroup$ I was mistaken. So the above table does indeed form a group since I checked all the four conditions? $\endgroup$ – speespa Feb 8 '15 at 22:13
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    $\begingroup$ If you've correctly checked the Group Axioms and your structure satisfies all of them, then it is a group! $\endgroup$ – mathmandan Feb 8 '15 at 22:15
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    $\begingroup$ With the translation $e=0$, $a=1$, $b=3$, $c=2$, our table is the addition table modulo $4$. $\endgroup$ – André Nicolas Feb 8 '15 at 22:33
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With the translation $e=0$, $a=1$, $b=3$, and $c=2$, we can recognize that our table is the addition table modulo $4$. More formally, the structure $M$ with the given multiplication table is isomorphic to the additive group $\mathbb{Z}_4$, via the mapping $\varphi$ that takes $e$ to $0$, $a$ to $1$, $b$ to $3$, and $c$ to $2$. The fact that the table is a group table then follows from the standard fact that $\mathbb{Z}_4$ is a group.

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