3
$\begingroup$

As referred in the Visual group theory Book by Nathan Carter- The unofficial definition of a group says that :

A group is a collection of actions satisfying the rules:
1. there is a predefined list of actions that never change.
2. Every actions is reversible.
3. Every actions is deterministic.
4. Any sequence of consecutive actions is also an action.

then the appearence of a Cayley's Diagram should be such that the above rules are satisfied.

Now the question is that the following figure satisfies the above rules but is still not a cayley's diagram:

enter image description here

The reason the book states is that it is so because the group lacks regularity i.e. it does not repeats every one of its internal pattern throughout the whole diagram.

I can't understand what role does 'regularity' in cayley's diagram has to do with condition for being a group.

$\endgroup$

2 Answers 2

2
$\begingroup$

A graph can be a Cayley diagram only if all the subgraphs consisting of blue edges (and the adjacent vertices) are isomorphic, and all the subgraphs containing red edges (and the adjacent vertices) are isomorphic (this generalises to more generators). This is essentially what "regularity" is getting at.

In this example, it is not regular because the top red region is different from the bottom two red regions. To utilise this to find a contradiction, note that the blue edges imply that there is a symmetry along them (obtained by swapping as follows: $1\leftrightarrow5$, $2\leftrightarrow6$, $3\leftrightarrow7$, $4\leftrightarrow 8$). However, this is not a symmetry of your graph!

$\endgroup$
2
  • $\begingroup$ can you please help explaining what is the reason behind : "A graph can be a Cayley diagram only if all the subgraphs consisting of blue edges (and the adjacent vertices) are isomorphic, and all the subgraphs containing red edges (and the adjacent vertices) are isomorphic " , as stated by you ...Kindly help me with this ... $\endgroup$
    – spectraa
    Jan 1, 2015 at 5:29
  • $\begingroup$ @spectra It is because of symmetry - the "blue" symmetry must preserve the "red" symmetry, but here when you flip along the blue edges you do not get the same graph back. $\endgroup$
    – user1729
    Jan 1, 2015 at 14:27
0
$\begingroup$

This is why the definition you've given from Visual Group Theory (excellent text!) is only an informal definition. Several of the statements in the definition are somewhat unclear in their meaning, which is why you want to move on to a more formal definition at some point. A way we can see that this diagram fails to be a Cayley diagram with direct reference to the rules of groups is to think of successive application of the "red" operation in different cases. If you apply red (r) twice to node 5 then you get back to node 5. This implies that rr = e, where e is the identity. However, if you apply r twice starting from node 1 then you end up at node 3, and so we conclude that rr is not e. I think in the informal definition this is violating the condition that actions are deterministic - i.e. that a given sequence of operations will always have the same result.

In terms of more formal definitions it is violating something that is implied but not usually stated in the definition of a group. There is a rule for combining group elements and the combination of two group elements is another group element (condition of closure). What is implied, but not usually stated here is that for any three elements satisfying ab = c, if they satisfy this relation some of the time then they must satisfy it all of the time. That is, there is an explicit set of unchanging relations relating the group elements to each other. This is part of the implied meaning by saying that "a rule for combining elements exists".

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .