What is the maximum order element in $S_{13}$ I want to find the maximum possible order of an element in $S_{13}$. I have found that disjoint cycles of length $(2)(3)(7)$ give order 42,
However ,answer in my book is given to be $60$
Can someone tell me how can I achieve given order ?
 A: Every element in $S_{13}$ can be written as a product of disjoint (meaning every number appears only once) cycles. Clearly, the order of a product of disjoint cycles (which, being disjoint, commute) is the least common multiple of the order of each cycle.
(clarifying example: $(12)(345)$ has order $6 = 2 \cdot 3$, because $(12)^2 =1$ and $(345)^3 = 1$) 
So to find the maximal order of an element in $S_{13}$ you need to find the partition of $13$ with the highest l.c.m. of the summands. Certainly $3+4+5(+1)$ gives you l.c.m. $60$.  The question is: are there bigger ones? I am afraid that "smart" trial and error is the way to go: ideally you want as many different primes as you can get, with the highest possible powers. Possible primes are $2,3,5,7,11,13$: easily, $11$ and $13$ are too big. $7$ would be a possibility, but the biggest l.c.m.s you can get with it are $2\cdot 3 \cdot 7 = 42$, or $5\cdot 7 = 12$, or $2 \cdot 4 \cdot 7 = 56$, all lower than $60$. So, after observing that $2^3 \cdot 5 = 40$, it has to be $3+4+5+1$.
(an interesting fact: in a somewhat counterintuitive way, the highest order in $S_{13}$ is achieved by an element of $S_{12}$!)
