Talk:Cayley's theorem

Latest comment: 2 years ago by Madyno in topic G finite?

How on the earth do people understand this?? — Preceding unsigned comment added by 24.77.20.249 (talkcontribs)

   First, you learn group theory. — Preceding unsigned comment added by 141.157.188.53 (talkcontribs) 

The wording throughout does not always distinguish carefully between G as a Group and G as a Set (i.e. U(G) or Forgetful(G)). I think that this distinction is critical to understanding this theorem since it is about a natural transformation between functors between categories Group and Set. —Preceding unsigned comment added by Tekolste (talkcontribs) 20:10, 12 July 2010 (UTC)Reply

Wrong comment

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  Thus, theorems which are true for permutation groups are true for groups in general.

That is ridiculous, the property "G is a permutation group" is true for any permutation group, but not true in general… (in the same spirit, "G is not finite, or there exists some k such that Card(G) = k!" does not hold for any group even if it holds for any permutation group). — Preceding unsigned comment added by Sedrikov (talkcontribs) 20:35, 18 June 2012 (UTC)Reply

LaTeX

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Many of the equations in this article can be produced in LaTeX. Some of the equations might be more readable if they were converted to LaTeX from plaintext. —Preceding unsigned comment added by 64.80.225.13 (talk) 17:18, 11 April 2011 (UTC)Reply

G finite?

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This may be too obvious, but probably should spell it out explicitly in the introduction. 130.91.186.4 (talk) 17:23, 12 May 2014 (UTC)Reply

G doesn't need to be finite - it can always be naturally embedded in its automorphism group in Set. Of course if G is infinite then that group isn't isomorphic to S_n for any finite n. 2001:BF8:200:FF68:2E41:38FF:FE93:5FFD (talk) — Preceding undated comment added 09:18, 2 July 2014 (UTC)Reply

But does the original form of the theorem requires the group to be finite? Sorry, forgot to login. Madyno (talk) 16:18, 14 November 2022 (UTC)Reply

History: Cayley's theorem and Dyck's theorem

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Our article says:

Burnside attributes the theorem to Jordan

and the reference given is the 1911 edition of Burnside's Theory of Groups of Finite Order, unfortunately with no page number. The 1897 edition of the same book calls it “Dyck's theorem”:

THEOREM: Every group of finite order   is capable of representation as a group of substitutions of   symbols.

(§20 (“Dyck’s Theorem”), p. 22.)

(“Substitutions”, in Burnside’s terminology, are what we now call permutations.)

The reference given is:

Dyck, “Grüppentheoretische Studien,” Math. Ann., Vol. xx (1882), p. 30.

“Dyck” here is Walther Dyck, later known as Walther von Dyck.

Clearly this doesn't predate Cayley's discovery of the theorem or Jordan's publication of it, but it might be worth mentioning anyway. Perhaps Burnside was unaware of Jordan's earlier publication of the theorem, it was brought to his attention after the first edition of his book, and he corrected the attribution in the second edition.

I am not sure what to do with it, but perhaps it ought to be mentioned? —Mark Dominus (talk) 14:48, 6 January 2017 (UTC)Reply

(Addendum: I cannot find an online copy of the second edition of the book. The Internet Archive copies are all the 1897 edition, despite being described as being from 1911. The Project Gutenberg version is 1897 also. —Mark Dominus (talk) 14:57, 6 January 2017 (UTC))Reply

There is a copy in Google Books. The page is again (§20, p. 22.) This time the theorem is stated this way:

THEOREM: Every group of finite order   can be represented as a group of regular permutations of   symbols.

The reference given this time is:

Jordan, Traité des Substitutions (1870), pp. 60, 61.

The title of section 20 has been changed from “Dyck's Theorem” to “Representation of a group of order   as a group of regular permutations of   symbols.”

— —Mark Dominus (talk) 15:04, 6 January 2017 (UTC)Reply