Make a list the elements of the subgroup  in S7 Consider the permutation $$a=\begin{pmatrix}1234567\\3276514\end{pmatrix}$$ in S7 .List the elements of the subgroup  in S7.
Hello, I just graduated from high school and in my summer time I'm reading about abstract algebra I want to learn more about the subject, but I do not understand how the book explains the part of permutation maybe I need a course, but I'll take it on University. Someone could explain or help me with this problem so I can understand better.
 A: The key step is writing the permutation into disjoint cycles:
$$
a=(1\,3\,7\,4\,6)(2)(5)
$$
See this answer of mine for a description.
Then the distinct powers of $a$ are
\begin{align}
a^0&=\text{identity} \\
a^1&=(1\,3\,7\,4\,6)(2)(5) \\
a^2&=(1\,7\,6\,3\,4)(2)(5) \\
a^3&=(1\,4\,3\,6\,7)(2)(5) \\
a^4&=(1\,6\,4\,7\,3)(2)(5)
\end{align}
If you want to write, say, $a^3$ in the “matrix form”, recall that each element has its image at its immediate right (bouncing back when the parenthesis is found):
$$
a^3=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
4 & 2 & 6 & 3 & 5 & 7 & 1
\end{pmatrix}
$$
A: The notation
$$a = \begin{pmatrix} b_1 & b_2 & \cdots & b_n \\ c_1 & c_2 & \cdots & c_n \end{pmatrix}$$
denotes a function with $a(b_1) = c_1, a(b_2) = c_2,\dots,a(b_n) = c_n$. This is usually given with $a_1 = 1, a_2 = 2, \dots, a_n = n$ but if you think about it, the order that the columns are given in doesn't matter. That is,
$$\begin{pmatrix}1&2&3&4&5&6&7\\3&2&7&6&5&1&4\end{pmatrix} \text{ and } \begin{pmatrix} 3&2&7&6&5&1&4\\7&2&4&1&5&3&6 \end{pmatrix}$$
denote the same function. This is helpful because $a^2(i) = a(a(i)) = i$ and we can easily compute this by stacking the two matrices on top of each other:
$$ \begin{pmatrix} \begin{pmatrix}1&2&3&4&5&6&7\\3&2&7&6&5&1&4\end{pmatrix} \\ \begin{pmatrix} 3&2&7&6&5&1&4\\7&2&4&1&5&3&6 \end{pmatrix} \end{pmatrix} \to \begin{pmatrix}1&2&3&4&5&6&7 \\7&2&4&1&5&3&6 \end{pmatrix} = a^2. $$
Notice that if you apply $a$ to $1$ you get $a(1) = 3$ and then if you apply $a$ again you get $a(a(1)) = a(3) = 7$ and the same works for $a(a(2)) = 2$ and $a(a(3)) = 4$ and so on.
Thus you can compute products with this stacking technique. Since the subgroup $\langle a \rangle$ generated by $a$ is nothing but the set of powers of $a$, this lets you compute the subgroup.
Note that this description of the subgroup generated by $a$ is only valid for a finite group. For an infinite group you need to add in $a^{-1}$ and its powers. The reason you can avoid this for finite groups is because in that case, $a^{-1}$ is always equal to a positive power of $a$.
