There's a math problem in Pinter's Abstract Algebra book on pp. 52(exercise G):

Exercise:

Every finite group may be represented by a diagram known as a Cayley diagram. A Cayley diagram consists of points

joined by arrows.

There is one point for every element of the group.

The arrows represent the result of multiplying by a generator.

For example, if $\displaystyle G$ has only one generator $\displaystyle a$ (that is, $\displaystyle G$ is the cyclic group $\displaystyle \langle a \rangle$), then the arrow $\displaystyle \rightarrow$

represents the operation "multiply by $\displaystyle a$":

$\displaystyle e\rightarrow a \rightarrow a^2 \rightarrow a^3 \rightarrow \cdots $

If the group has two generators, say $\displaystyle a$ and $\displaystyle b$, we need two kinds of arrows, say dotted arrow and

line arrow with no dots where dotted arrow means "multiply by $\displaystyle a$" and lined arrow means "multiply by $\displaystyle b$".

For example, the group $\displaystyle G = \{ e, a, b, b^2, ab, ab^2 \}$ where $\displaystyle a^2 = e, b^3 = e, $ and $\displaystyle ba = ab^2$ has the following

Cayley diagram(Figure 1):

Attachment 25767

Moving in the forward direction of the lined arrow means multiplying by $\displaystyle b$,

$\displaystyle x \rightarrow xb$

whereas moving in the backward direction of the lined arrow means multiplying by $\displaystyle b^{-1}$:

$\displaystyle x \leftarrow xb^{-1}$

(Note that "multiplying $\displaystyle x$ by $\displaystyle b$" understood to mean multiplying on the right by $\displaystyle b$:

it means $\displaystyle xb$, not $\displaystyle bx$) It is also a convention that if $\displaystyle a^2 = e$(hence $\displaystyle a = a^{-1}$), then no

arrowhead is used:

$\displaystyle x\;\;............\;\;xa$

for if $\displaystyle a = a^{-1}$, then multiplying by $\displaystyle a$ is the same as multiplying by $\displaystyle a^{-1}$

The Cayley diagram of a group contains the same information as the group's table. For instance, to find the product $\displaystyle (ab)(ab^2)$

in the previous figure 1, we start at $\displaystyle ab$ and follow the path corresponding to $\displaystyle ab^2$ (multiplying by $\displaystyle a$, then

by $\displaystyle b$, then again by $\displaystyle b$), which is (Figure 2)

Attachment 25766

This path leads to $\displaystyle b$, hence $\displaystyle (ab)(ab^2) = b$

As another example, the inverse of $\displaystyle ab^2$ is the path which leads from $\displaystyle ab^2$ back to $\displaystyle e$.

We note instantly that this is $\displaystyle ba$.

A point-and-arrow diagram is the Cayley diagram of a group iff it has the following two properties:

1) For each point $\displaystyle x$ and generator $\displaystyle a$, there is exactly one $\displaystyle a$-arrow ending at $\displaystyle x$;

furthermore, at most one arrow goes from $\displaystyle x$ to another point $\displaystyle y$.

2) If two different paths starting at $\displaystyle x$ lead to the same destination, then those two paths, starting at any point $\displaystyle y$, lead

to the same destination.

Cayley diagrams are a useful way of finding new groups.

Problem:

Write the table of the groups having the following Cayley diagram (Remark: You may take any point to represent $\displaystyle e$(neutral element), because

there is perfect symmetry in a Cayley diagram. Choose $\displaystyle e$, then label the diagram and proceed.)

Attachment 25763

My problem is: Is it possible to kindly help me find this cyclic group's operation table? How do I choose the elements from this diagram?