Subgroup of a symmetric group $S_7$ Is there any quick way to determine if $S_7$ contains a subgroup of order $6$ or $S_{11}$ a subgroup of order $30$ ? A problem carrying only 2 marks involves this. So I assume either there is some quick way of finding it or a prior knowledge is required. Your feed back please.
 A: What are the groups of order six? These are $S_3$ and $C_6$. Is there a copy of $S_3$ in $S_7$? Yes. Is there a copy of $C_6$ in $S_7$? Equivalently, is there an element of order $6$? Yes (e.g. any $6$-cycle).
What are groups of order $30=2\cdot3\cdot5$? Most obvious would be $S_3\times C_5$ or $C_{30}\cong C_2\times C_3\times C_5$; are these available in $S_{11}$? Yes: for the first, take an obvious copy of $S_3$, then take a $5$-cycle on any other five elements not touched by $S_3$. What about $C_{30}$?  Obviously there's no $30$-cycle. Knowledge of how cycle types correspond to orders tells us it suffices to find permutation of type $(2,3,5)$ i.e. a $2$-cycle, $3$-cycle, and $5$-cycle all disjoint. This is certainly possible, as $2+3+5=10\le11$. One could also take a disjoint $5$-cycle and $6$-cycle as $5+6\le11$.
A: A quick way to determine a subgroup of order 6 in $S_7$ is to find an element of order 6. That can be done by taking a set of positive integers whose sum is 7 and lcm is 6. For example take $\{2,2,3\}$. Then $(1,2)(3,4)(5,6,7)$ has order 6. For $S_{11}$, the element $(1,2,3,4,5)(6,7,8,9,10,11)$ has order 30.
A: More generally, $S_n$ contains a group of order $m$ if $m$ has the form $m=a_1\cdot a_2\cdots a_k$ where $n\geq a_1+a_2+\ldots+a_k$. If $m$ has this form then simply take some relevant disjoint $a_i$-cycles (the resulting subgroup will be $C_{a_1}\times C_{a_2}\times\cdots\times C_{a_k}$).
For example, $6=2\cdot 3$ and $2+3=5$ so take $(1, 2)$ and $(3, 4, 5)$. Thus, $S_5$ contains a subgroup of order six (and thus so does $S_7$). This subgroup is $C_2\times C_2\cong C_6$. The second example also works like this because $30=2\cdot3\cdot5$ and $2+3+5=10<11$.
