Set of permutations of $14$ elements in $S_{14}$ How do you know this set of permutation is not cyclic? I know there are $14!$ elements in the group, but I don't know how to prove that $S_{14}$ is not cyclic.
 A: Explaination
Well the group generated by all permutations on the set $\{1,...,n\}$ ($S_n$) is non-abelian (a stronger result) for $n\geq 3$. First show that this is true for $S_3=\{e,(1,2),(1,3),(2,3),(1,2,3),(1,3,2)\}$. Simply find an 2 elements that don't commute. try $(1,2),(1,3)$.
Then show that $S_3$ is a subgroup of $S_n$ for $n>3$, (you can just note that the set of permutation fixing $\{4,...,n\}$ and permuting $\{1,2,3\}$ is a subgroup of $S_n$).
Possible questions

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*Wait what does $(1,2,3)$ mean?
Answer: It means that $1\to 2$, $2\to 3$ and $3\to 1$, also $(1,2)$ means $1\to 2$, $2\to 1$ and $3\to 3$


*Wait why does $S_3=\{e,(1,2),(1,3),(2,3),(1,2,3),(1,3,2)\}$?
Answer: This is a bit complex, but essentially if you have decide what kind of permutation you are going to have say $f(x)=\begin{cases}1 & x=2\\ 2 & x=1\\ 3 & x=3\end{cases}$ then you can go down the list and go along the cycles for ($1\to 2\to 1$) and just rewrite it into a "product" of cycles
(rewrite $(1\to 2\to 1)$ as $(1,2)$ and $3\to 3\to 3$ as $(3)$ which is often times left out . So we have $(1,2)(3)=(1,2)$).


*Why start with abelian and not just go straight to proving it isn't cyclic
Answer: It is a more powerful result that I feel you would have ran into trying to prove that $S_14$ isn't cyclic.


*How does $S_{14}$ not being abelian show that it isn't cyclic
Answer: A cyclic group is abelian the converse statement isn't true.


*Wait, what do you mean by " the set of permutations fixing $\{4,...,n\}$ and permuting $\{1,2,3\}$ is a subgroup of $S_n$"
Answer: I am saying take all function of the form $f(x)=\begin{cases}1 & x=a\\ 2 & x=b\\ 3 & x=c\\ 4 & x=4\\...\\ n & x=n\end{cases}$ where $a,b,c\in\{1,2,3\}$ and ($a,b$, and $c$ are distinct), and put them in a set. Then you can check that this is essentially $S_3$.
A: Cyclic groups of order $n$ have exactly one subgroup of order $d$, for each $d$ dividing $n$.
Now, $2$ divides $14!$, and I bet you can find at least two subgroups of $S_{14}$ that have order $2$.
