Find the subgroup of $S_4$ generated by $(1,2,4)$. What is the order of $(1,2,4)$? I am literally just being introduced to group theory this is all new. I know that the permutations that arise are $(4,1),(1,2),(2,4),(1,2,4)$ This is the same as saying that $4$ goes to $1$ then $1$ goes to $2$ and then $2$ goes to $4$ and then we have the trivial case which is itself $(1,2,3,4)$ is this what I am supposed to show? Isn't the order just the number of subgroups?
 A: $(124)$ means $1\to2\to4\to1$.  It's a $3$-cycle.  As such, it generates a cyclic group of order $3$. An $n$-cycle has order $n$.  So, it generates a cyclic subgroup of order $n$.
The subgroup is $\{e,(124),(142)\}$.
A: I would like to explain very simply.
$$\sigma=(1,2,4)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)$$$$$$then we must calculate $\sigma, \sigma^2,...$ until we reach $$e=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
1 &2&3&4 \end{array} \right)=(1) $$$$not $$$$(1,2,3,4)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &3&4&1 \end{array} \right)$$$$$$ $$\sigma=(1,2,4) , \sigma^2=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
4 &1&3&2 \end{array} \right)=(1,4,2)$$$$ $$ $$\sigma^3=\sigma.\sigma^2=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
4 &1&3&2 \end{array} \right)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
1 &2&3&4 \end{array} \right)=(1)       $$ $$ $$ so the subgroup generated with $\sigma=(1,2,4)$ is:  $$$$ $$\left\{ \begin{array}{c}\sigma=(1,2,4),\sigma^2=(1,4,2),\sigma^3=(1)\end{array}\right\}.$$ $$$$ the order of $(1,2,4)$ is the number of elements of subgroup generated by  $(1,2,4)$ so $O(\sigma)=3$
