# What is the largest abelian subgroup in $S_n$?

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.

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).
• 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. – Derek Holt Dec 4 '15 at 12:21