Talk:Cayley table

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Clarification of Commutativity subsection[edit]

The sentence: "The Cayley table tells us whether a group is abelian. Because if the group operation of an abelian group is commutative, the Cayley table of an abelian group is symmetric along its diagonal axis." does not seem totally clear, for two reasons: first, every abelian group is commutative (so the condition "the group operation of an abelian group is commutative" is always fulfilled) and second it could (should?) be stated as an equivalence : "The Cayley table tells us whether a group is abelian or not: the group operation is commutative if and only if the Cayley table of the group is symmetric along its diagonal axis." Don't you think so? 164.81.50.98 (talk) 09:00, 22 September 2009 (UTC)

Request for simple example[edit]

Perhaps someone should add a simple example to this page like

   + 0 1
   0 0 1
   1 1 0

I might do it later but in the meantime, back to work .... OoberMick 11:04, 9 Jun 2005 (UTC)

Caryler table applet[edit]

I created a Cayley Table generator and identifier applet, but I do not know if it is appropriate to add an external link to it. Jhbarr 16:46, 18 October 2007 (UTC)

Cayley table reasoning[edit]

I don't understand the reasoning at the end of the article's long example: "As we have managed to fill in the whole table without obtaining a contradiction, we have found a group of order 6". I agree that the structure obtained is a group of order 6, but I don't enough work has been done in the example to deduce that it's a group, because associativity was not checked. Or did I miss something? —Preceding unsigned comment added by 134.193.88.39 (talk) 15:56, 11 May 2008 (UTC)

I agree. It does not appear to me that enough conditions have been checked. A very similar argument in structure is coset enumeration, and there if a "cayley" table is filled in (at least the columns corresponding to generators, and the rows corresponding to coset representatives of a subgroup H), then the order of the group (or at least the index of H in G), is found. Using H=1, this does allow you to fill in a few columns of the Cayley table from a presentation of the group. JackSchmidt (talk) 16:09, 11 May 2008 (UTC)
I guess the question then is, is it possible to follow this algorithm and obtain a Cayley table for a quasigroup that is not also a group? Most of the algebraic manipulations used here require associativity, but it's true that an exhaustive check has not been performed. It's an interesting question... 72.42.181.162 (talk) 17:05, 13 July 2008 (UTC)

paragraph Filling in the identity skeleton[edit]

I dont follow the reasonning in the grid where 'e' is marked in d*f box instead of d* d box etc. I mean if e is the identity element ? Hexagone59 (talk) 20:50, 25 December 2015 (UTC)

Article in need of cleanup[edit]

The article, as it stands, seems to be geared towards someone who already has been taught what a Cayley table is and how to construct one.

Structure and layout[edit]

Understanding the first table is easy with the included text describing the row,col order. The second table, and it's description make broad assumptions about the reader's knowledge. I can see, for example, the cyclic nature of the second table, but there's no description as to why it's set up that way, or if it's even related to the first table. The addition of "It is easy to see..." before showing a vague statement like "b2 = c", which makes no real sense and is definitely not "easy to see". I can see that if you treat the two 'b's as row and column headers, that 'c' would be the result, but why that is the case makes no sense and that is the end of that example.

Constructing Cayley tables[edit]

This section falls apart immediately.

  • "By convention, e is the group's identity element."...so why was 'a' used above?
  • "Because the identity element is always its own inverse, and inverses are unique, [then?] the fact that there are 6 elements in this group means that at least one element other than e must be its own inverse."...why?
  • "the fact that there are 6 elements in this group means that at least one element other than e must be its own inverse"...why?
  • "So we have the following possible skeletons:"
  • all elements are their own inverses,
  • all elements save d and f are their own inverses, each of these latter two being the other's inverse,
  • a is its own inverse, b and c are inverses, and d and f are inverses. why?
Are these three possibilities for a skeleton or are they all part of the same skeleton as it seems in the example that follows?
  • Why is the initial table all 'e's? This doesn't conform to the Structure and layout section above.
  • "Obviously, the e row and the e column can be filled out immediately."...why would this be obvious?
  • "...it may be necessary (and it is necessary, in our case) to make an assumption..."...what makes it necessary?
  • "...which may later lead to a contradiction..."...what's a contradiction? Are we going to see one later?
  • "Multiplying ab = c on the left by a gives b = ac..."...on the left of what? Why do the 'a's cancel?
  • "...Multiplying on the right by c gives bc = a."...which equation? on the right of what? Are these steps on the initial assumption? That's what it looks like...but then "on the left/right" makes no sense. What part is the result? Where are we putting the result? On that table I haven't scrolled to yet? Why are those letters where they are on that table.

From here, I glanced through the rest and it just confounded me and I gave up. Someone with knowledge on the subject please clean this up. —Zephalis (talk) 00:41, 5 January 2016 (UTC)