The following is a simple exercise from "Guide to Abstract Algebra 1st Ed Carol Whitehead".
Problem 6.2b.1
The following is part of a Cayley table for a group $G = {a,b,c,d}$, with respect to operation $\circ$.
$$ \begin{array}{l|*{4}{lr}} \circ & a & b & c & d \\ \hline a & c & \cdot & \cdot & \cdot \\ b & \cdot & \cdot & \cdot & a \\ c & \cdot & \cdot & \cdot & d \\ d & \cdot & \cdot & \cdot & \cdot \\ \end{array} $$
(a) Which is the identity element of $(G,\circ)$?
(b) Show that there is only one way to fill the rest o the table, justifying each step. Note: You are not required to verify that $\circ$ is associative.
My Solution Attempt:
(a) We can see $c \circ d =d$, so the identity element is $c$.
(b) Since $c$ is the identity, we can fill in the $c$ rows and columns.
$$ \begin{array}{l|*{4}{lr}} \circ & a & b & c & d \\ \hline a & c & \cdot & a & \cdot \\ b & \cdot & \cdot & b & a \\ c & a & b & c & d \\ d & \cdot & \cdot & d & \cdot \\ \end{array} $$
I know also from an earlier theorem in the book that each element of a group appears only once in each row or column.
Looking at $a \circ b$, this could be $b$ or $d$ if we note that $c$ and $a$ are already in that row. But $b$ is already in that column so $a \circ b$ must only be $d$.
Following this logic gives us the following:
$$ \begin{array}{l|*{4}{lr}} \circ & a & b & c & d \\ \hline a & c & d & a & b \\ b & d & c & b & a \\ c & a & b & c & d \\ d & b & a & d & c \\ \end{array} $$
My Question: Is this correct?
I notice that in Cayley tables for groups, the rows and columns always follow a single order, or its reverse. That is $a,b,c,d$ or $d, c, b, a$ but never something like $a, d, c, b$. Additional Question: What is the reason for this?