As an important tool for visualizing some small finite groups it is useful to know how read such graph, and with time trying to make sketch of them by my own.

I would like to know, for a start, how to read properly the cycle graph of the symmetric group of order 3?

enter image description here

  • 2
    It might be worth mentioning that the author of this diagram is notorious in the Wikipedia math community for assembling many complex, attractive diagrams that nobody else can easily understand. – MJD Apr 11 '15 at 15:09
  • @MJD Who are you talking about? – Santropedro Jan 18 '17 at 0:30
up vote 4 down vote accepted

The group $S_3$ is generated by two elements, each of order 2, which we may call $a$ and $b$.

The central node in the diagram represents the identity element. For each of the other 5 elements, the diagram contains one node and one cycle. The cycle for an element of order $n$ contains exactly $n$ nodes, including the identity element node.

The cycle for an element $x$ of order $n$ consists of the nodes for elements $e, x, x^2, x^3,\ldots x^{n-1}$.

Two of the elements of $S_3$ have order 3. (These are $ab $ and $ba $.) You can see their shared 3-cycle in the top half of the diagram. The cycle for element $ab $ consists of nodes $e, ab, (ab)^2=ba $ after which it returns to $e$ because $(ab)^3=e$. The cycle for $ba $ is exactly the same but in reverse order because $ ba=(ab)^{-1} $.

Three of the elements in your example have order 2. ($a, b, $ and $aba=bab $). Each of these corresponds to one of the leaves at the bottom of the figure. For example the lower left node represents the cycle for $ a $, which consists of nodes $ e, a $, and then returns to $ e $ because $ a^2=e $. Properly speaking these leaves should have one edge on each direction, but we abbreviate the two edges to one to avoid clutter.

The other stuff in this diagram is nonstandard stuff that the somewhat eccentric author made up. Don't worry if it's not clear; this author is notorious for that.

  • .@MJD This description is what I was on the verge of posting and is correct. I can add that this is not a Cayley graph w.r.t. the generating set $\{a,b\}$ because there is no edge between $ab$ and $aba$. – Duchamp Gérard H. E. Apr 11 '15 at 16:08
  • Thanks for pointing that out. – MJD Apr 11 '15 at 16:59

I don't think I can answer everything that is in this graph, but we can make a dent. First of all, since this is the symmetric group, if we interpret matrix entry $m_{1,1}$ as element 1 of a set, matrix entry $m_{2,2}$ as a second element of a set and so on, then the arrows show the group acting on that set. So there are no arrows on the identity, which acts trivially on the elements. The red and white squares show a 3-dimensional representation of the group. Furthermore, the levels in the graph (2 on top, one in the middle, 3 on bottom) come from the order of the different elements.

I cannot answer why the elements were numbered the way they were, or why some of the numbers are green instead of grey.

  • The numbering is probably arbitrary. The green/grey probably means parity. – Rufflewind Aug 22 '15 at 0:15

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