# How to show that this Cayley Table does not form a group

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

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.

• It's surprisingly difficult to prove things that are false! – Derek Holt Feb 8 '15 at 22:05
• "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? – mathmandan Feb 8 '15 at 22:11
• I was mistaken. So the above table does indeed form a group since I checked all the four conditions? – speespa Feb 8 '15 at 22:13
• If you've correctly checked the Group Axioms and your structure satisfies all of them, then it is a group! – mathmandan Feb 8 '15 at 22:15
• With the translation $e=0$, $a=1$, $b=3$, $c=2$, our table is the addition table modulo $4$. – André Nicolas Feb 8 '15 at 22:33

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.