# A basic mistake by Cayley

Arthur Cayley's first paper on abstract groups, in 1854, can be found in his Collected Papers on the Internet Archive, starting at https://archive.org/stream/collectedmathema02cayluoft#page/122/mode/2up. He states on the 3rd page that groups of prime order are cyclic and then goes on to classify the groups of order 4 and 6 up to isomorphism (he doesn't use that term and refers to the order of an element as its "index," but to modern eyes it is clear what he is doing). If you go to the 5th page of the paper, at https://archive.org/stream/collectedmathema02cayluoft#page/126/mode/2up, he is classifying groups of order 6 and writes near the bottom "we have thus two, and only two, essentially distinct forms of a group of six [symbols]." The next page shows the Cayley tables for these groups, corresponding to the cyclic group and $S_3$, respectively. So far so good.

Moving ahead to 1878, in the first volume of the American Journal of Mathematics Cayley has a short paper on groups. See https://www.jstor.org/stable/2369433?seq=1#page_scan_tab_contents. On the bottom of the second page he writes "The general problem is to find all the groups of a given order $n$." He goes on to say that when $n = 2$ and $3$ there is one group, when $n = 4$ there are two groups (a footnote points out when $n = 5$ there is one group), and then comes the big surprise: "when $n = 6$, there are three groups". That is not only wrong, but contradicts what he wrote in 1854.

Cayley describes the three groups of order 6 as $1, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5$ where $\alpha$ has order 6; and two groups of the form $1, \beta, \beta^2, \alpha, \alpha\beta, \alpha\beta^2$ where $\alpha$ has order 2 and $\beta$ has order 3, distinguished from each other by the rules $\alpha\beta = \beta\alpha$ or $\alpha\beta = \beta^2\alpha$. (Strictly speaking, he lists that last relation together with $\alpha\beta^2 = \beta\alpha$, but it's redundant since it follows from $\alpha\beta = \beta^2\alpha$ since $\beta^3 = 1$.) The second group in his list is recognizable as a direct product of a group of order 2 and a group of order 3, so it appears that Cayley does not realize his second group is isomorphic to his first group. The mistake is understandable; even today, students learning group theory who do not yet know the Chinese remainder theorem could easily believe at first that a direct product of cyclic groups of order 2 and 3 is not isomorphic to a cyclic group of order 6 since the groups don't "look" the same (confusing appearance with structure).

My question is this: did Cayley in print later acknowledge his mistake or explain why he got it right in 1854 and later got it wrong? In a 1913 Amer. Math. Monthly paper G. A. Miller writes about Cayley's mistake on https://www.jstor.org/stable/2973510?seq=4#page_scan_tab_contents, but is silent about Cayley ever issuing a correction or explanation later.

Note: Near the end of writing up this question I found Cayley's error mentioned by Richard Stanley in a Mathoverflow post about mathematical errors -- see https://mathoverflow.net/questions/879/most-interesting-mathematics-mistake/921 -- but my question still stands: did Cayley ever bring up his error after 1878?

• No comment is made on the paper’s 1896 reprint in Cayley’s collected works. (As long as he edited those himself, volumes ended with a “Notes and References” section for such comments, but he didn’t live to edit this one. The index volume lists his later paper on groups.) – Francois Ziegler Nov 12 '17 at 21:09
• @FrancoisZiegler in the copy of his collected papers that you link to (scanned from Berkeley's library) I see someone underlined the "three" and wrote a question mark below it. – KCd Nov 12 '17 at 21:19
• Yes! Unless I am mistaken, he repeated the mistake in a 1891 paper (Collected Papers XIII, p. 125). Is it possible that he sometimes meant to classify not up to isomorphism but some other relation (e.g., conjugacy within the symmetric group on |G| letters, or something like that)? – Francois Ziegler Nov 12 '17 at 21:47
• @FrancoisZiegler, on p. 118 he has a table that lists the number of "groups" in $S_n$ for small values of $n$. For $n = 4$ he lists the count 7, but there are 11 conjugacy classes of subgroups: he doesn't count the trivial subgroup but missed one conjugacy class of order 2 and also subgroups of order 3 and 6. – KCd Nov 12 '17 at 23:32
• I don't think there is a charitable interpretation for his mistake in 1878: groups there are abstract, not inside of symmetric groups. – KCd Nov 12 '17 at 23:39