# Cayley table help

I have a Cayley table with four elements and a binary structure $*$. I know that if I have the same element along the main diagonal (from top right corner to bottom left corner), then the set is Abelian.

What can I say about the set if the table also has the identity element of the set going down the other diagonal (from the top left to the bottom right)?

Every element appears once in every row and column so I'm tempted to say it's a group. But say the set is {a,b,c,d} where a is the identity element so I need to find the inverse of every element to show it's a group. So, $a*a=a$, $b*b=a$, $c*c=a$, and $d*d=a$. How can an element operated on itself be an inverse?

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For the resulting group to be abelian, the matrix has to be symmetric: the entry corresponding to $a*b$ has to be the same as the entry corresponding to $b*a$. It does not require the same entry in the main diagonal. – Arturo Magidin Mar 5 '12 at 16:37
Okay, so it has to be symmetric. But what if it is also symmetric along the other diagonal and the identity element is the only element along that diagonal? Does this have any significance to determining the structure (say defining it as a group or anything) – user23793 Mar 5 '12 at 16:41
Note: The "main diagonal" is the name usually given to the diagonal running from top left to bottom right; the diagonal running from top right to bottom left is often called the "main anti-diagonal". – Arturo Magidin Mar 5 '12 at 16:42
@user23793: "Symmetric along the other diagonal" does not make sense. "Symmetric" means that it equals its transpose: the $(i,j)$ component equals the $(j,i)$ component for every $i$ and $j$. This is exactly "symmetric about the diagonal running from top left to bottom right." – Arturo Magidin Mar 5 '12 at 16:45
What I'm trying to say is that if the Cayley table reads "a b c d" along the top and "a b c d" along the side of the table and I have the identity element "a" in the spot where aa, bb, cc, and dd are and every element appears once in every row and column, does this mean the set is a group? The reason I am getting confused is because for example b*b=a doesn't seem like it makes sense but if it's perfectly fine to have this operation, then that's what I need to know. – user23793 Mar 5 '12 at 16:56

The "main diagonal" of a table/matrix refers to the entries with index $(i,i)$; this is the diagonal that runs from "top left" to "bottom right": it's the entries marked with a $\mathbf{D}$ below: $$\begin{array}{|c|c|c|c|} \hline \mathbf{D}& \cdot & \cdot & \cdot\\ \hline \cdot & \mathbf{D} & \cdot & \cdot\\ \hline \cdot & \cdot & \mathbf{D} & \cdot\\ \hline \cdot & \cdot & \cdot & \mathbf{D}\\ \hline \end{array}$$

If you have a Cayley table, and the elements along the top are ordered the same way as the elements along the side (so that the main diagonal entries correspond to $a*a$ for every $a$, and you already know that this is the Cayley table for a group, and every main diagonal entry is equal, then that entry must be the identity (since $e*e=e$ holds, so every entry must be $e$).

A group in which every element is its own inverse must be abelian: if $xx=e$ for every $x$, and $a$ and $b$ are any two elements, then we have that $(a*b)^2 = e = e*e = a^2*b^2$. So then we have $$a*b*a*b = a*a*b*b$$ and multiplying on the left by $a$ and on the right by $b$ we get $$b*a = a*b,$$ so the group is abelian.

To see an example of a group where this happens, consider the following operation: flipping a rectangular mattress. You can flip it end-to-end; you can rotate it without changing what is on top; you can flip it and rotate it; or you can do nothing. Each of these is an element of a group; if you do the same thing twice in succession, they cancel out. So every element of the group is its own inverse. There is absolutely no problem with that: there are lots of groups like that.

Here is what they look like:

Let us use $E$ to denote "Even", and $D$ to denote "Odd". We make a group with underlying set $\{E,D\}$, and add as follows: $E+E = E$; $E+D=D+E=D$; and $D+D=E$. (Even plus even and odd plus odd are both even; even plus odd and odd plus even or both odd).

Now let $n\geq 1$, and let $G$ be the set of all $n$-tuples $(x_1,x_2,\ldots,x_n)$, where $x_i$ is either $E$ or $D$. We add tuples by adding component by component: $$(x_1,x_2,\ldots,x_n) + (y_1,y_2,\ldots,y_n) = (x_1+y_1,x_2+y_2,\ldots,x_n+y_n),$$ where in each component we are adding following the rules of the previous paragraph. This is a group (it has $2^n$ elements); the identity element of the group is the element $(E,E,E,\ldots,E)$. And in this group, every element is its own inverse: $(x_1,\ldots,x_n) + (x_1,\ldots,x_n) = (E,E,E,\ldots,E)$, no matter what $x_i$ is: if $x_i=D$, then $x_i+x_i = D+D=E$; if $x_i=E$, then $x_i+x_i = E+E=E$. So either way, we get the identity.

However, if you simply happen to find a "Cayley table" on the ground in which every main diagonal entry is equal, this does not suffice to tell you that you have an abelian group before you: associativity is not easy to determine from just staring at the table, and it is possible to write down the Cayley table of a binary operation in which (i) every element appears exactly one in each row and in each column; (ii) the rows correspond to elements in the same order as the columns do; (iii) all the main diagonal entries are the same; but (iv) the table does not correspond to an associative operation (that is, you don't have a group).

Here's an example where this happens: $$\begin{array}{cc|c|c|c|} * && a & b & c\\ &&&&\\ \hline a && a & b & c\\ \hline b && c & a & b\\ \hline c && b & c & a\\ \hline \end{array}$$ Every element appears exactly once in each row and each column, but this table does not yield a group, because the operation $*$ it defines is not associative: $(b*c)*c = b*c = b$, but $b*(c*c) = b*a = c$.

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