Computing $\langle (13746) \rangle$ in $S_7$. How to list the elements of subgroup $\langle a \rangle$ in $S^7$ where $$a=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 3 & 2 & 7 & 6 & 5 & 1 &4 \end{pmatrix}?$$  I got $a=(13746)(2)(5)$.
Any help is appreciated!
 A: First $1$ goes to $3$, then it goes to $7$, then it goes to $4$, then it goes to $6$, then it goes to $1$.  In cycle notation that's $(13746)$.  $2$ and $5$ are fixed, so there's no need to think about them.
So, let's calculate $a^2$.  $1$ skips over $3$ and goes to $7$, which skips over $4$ to get to $6$, which skips over $1$ to get to $3$, which skips over $7$ to get to $4$, then back to $1$.  So $a^2=(17634)$.
In the same way, we get that $a^3=(14367)$.
Now you try for $a^4$.
Then, I claim that $a^5$ is the identity. Why?
That brings us to $\langle a \rangle = \{\operatorname{id},(13746),(17634),(14367),a^4\}$ (where you calculate $a^4$).
A: Your calculation of $a$ is correct, but (2) and (5) need not be written out. To find the elements of the subgroup, first calculate $a^2 = (1 7 6 3 4)$, then $a^3$, $a^4$, and so on until you get the identity element (1). There won't be very many calculations...
A: We have that
$
a=
\left(
\begin{array}{cccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 3 & 2 & 7 & 6 & 5 & 1 & 4
\end{array}
\right)
$, which can be expressed as $a=(13746)\\$.
Then $a^2=(13746)(13746)=(17634)\\$.
Similarly, $a^3=(17634)(17634)=(16473)$. We see that the $n^{th}$ power of $a$ takes the entry in the $i^{th}$ position of $a$ to the entry in $i+1^{th}$ entry of $a$. 
Since $a^5$ brings each entry of $a$ back to its original location, it is the identity, and $|a|=|⟨a⟩|=5$.
