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I am almost to complete my first course in group theory. I have read Dummit and Foote into chapter 5. I know that I can always find an abelian subgroup isomorphic to $C_{k_1} \times C_{k_2} \times ... \times C_{k_t}$ in $S_n$ where $n = k_1 + k_2 + ... + k_t$. Specifically, the permutation group $G = <(1 \ 2\ ...\ k_1),(k_1+1 \ ...\ k_2),...,(n-k_t+1\ ...\ n)>$.

There are many ways to express an isomorphism type of an abelian group as a direct product of cyclic factors. If I express the isomorphism type in its elementary divisor (primary) decomposition this seems to minimize $n$. I am somewhat familiar with Landau's function (Sloane's A000793) which gives the maximal lcm over all the partitions of $n$. I think that this sequence gives the answer to my question but there are no comments in the data base regarding this. At any rate, I am unsure and would like some help on this question.

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Since all direct factors contribute, you don't need the lcm of the orders. Also note that $|C_2\times C_2\times\cdots\times C_2|$ ($k$ factors, on $2k$ points) has order $2^k\ge 2k$, so it is beneficial to split any cycle of length $>3$ into groups of $2$-cycles. I fact (as Derek Holt observed - $2^3<3^2$ on 6 points) splitting into 3-cycles gives an even larger order, but going to larger cycle lengths will produce worse results. This means that the largest subgroup is the direct product of copies of $C_3$ with $2$-cycles thrown in on the remaining points (i.e. if $n\equiv 0\pmod{3}$ there is no $2$-cycle, if it is $\equiv 2$ there is $1$ two cycle, and if it is $\equiv 1$ we remove one 3-cycle and write two $2$-cycles (or a $V_4$, or a 4-cycle, as $4=2^2$) on the 3 points plus the remaining point).

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  • $\begingroup$ OK Thanks. I see now that we don't need to consider the lcm of the orders (of course not, the order of the direct product group is the PRODUCT of the orders of its factors). I computed (using Mathematica) the maximal product over the partitions of n to find Sloane's A000792. A comment in that sequence states that this is indeed the answer to my question. However this does not agree with your reply. Consider S_6, which contains the subgroup of order 9, <(1 2 3) , (4 5 6)>. $\endgroup$ – Geoffrey Critzer Dec 4 '15 at 10:51
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    $\begingroup$ Yes, $3^{n/3} > 2^{n/2}$ for large $n$, so you need to take mainly $3$-cycles, and just $1$ or $2$ $2$-cycles. $\endgroup$ – Derek Holt Dec 4 '15 at 12:21
  • $\begingroup$ We could then ask how many distinct abelian subgroups of maximal order are contained in S_n. I think these are enumerated by the exponential generating function, exp(x^3/3)*( 1 + exp(x^2/2 + x^2/2) + exp(x^4/4) ). $\endgroup$ – Geoffrey Critzer Dec 4 '15 at 14:40
  • $\begingroup$ The generating function that I gave above is NOT correct. $\endgroup$ – Geoffrey Critzer Dec 5 '15 at 14:24
  • $\begingroup$ I think that the number of abelian subgroups of maximal order in S_n for n=1,2,... is 1,1,1,6,10,10,210,280,280,12600,15400,... and is given by the e.g.f. exp(x^3/3!)*(1 + x^2/2 + x^4/4) + x - 1. This sequence is not given in OEIS. Can GAP verify these terms? $\endgroup$ – Geoffrey Critzer Dec 6 '15 at 11:40

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