Is the set of permutations in $S_{36}$ that move at most 4 elements a subgroup of $S_{36}$? I am truly lost as to what this problem is asking.  I did post this on another forum and received what my have been wonderful advice.  However, even after multiple hours and many "Google" searches I am still confused.  Please anyone....
The problem states:
If $\sigma$ is a permutation of a set $A$, we say that $\sigma$ moves $a\in A$ iff $\sigma(a)\neq a$.
For the symmetric group $S_{36}$ of all permutations of 36 elements, let $H$ be a subset of $S_{36}$ containing all permutations that move no more than four elements.  Is $H$ a subgroup of $S_{36}$?  Prove.
I am not looking for an answer but more an explanation of what the problem is saying.  I do not know what "move no more than four elements" means. I am feeling really stupid.  Any help appreciated.   
 A: Let's look at $S_3$ instead.  There are six permutations:  $\left(\begin{array}{ccc}
1 & 2 & 3  \\
1 & 2 & 3
\end{array}\right)$, $\left(\begin{array}{ccc}
1 & 2 & 3  \\
1 & 3 & 2
\end{array}\right)$, $\left(\begin{array}{ccc}
1 & 2 & 3  \\
2 & 1 & 3
\end{array}\right)$, $\left(\begin{array}{ccc}
1 & 2 & 3  \\
2 & 3 & 1
\end{array}\right)$, $\left(\begin{array}{ccc}
1 & 2 & 3  \\
3 & 1 & 2
\end{array}\right)$, and $\left(\begin{array}{ccc}
1 & 2 & 3  \\
3 & 2 & 1
\end{array}\right)$.  
$\left(\begin{array}{ccc}
1 & 2 & 3  \\
1 & 2 & 3
\end{array}\right)$ moves no elements.  $\left(\begin{array}{ccc}
1 & 2 & 3  \\
1 & 3 & 2
\end{array}\right)$, $\left(\begin{array}{ccc}
1 & 2 & 3  \\
3 & 2 & 1
\end{array}\right)$, and $\left(\begin{array}{ccc}
1 & 2 & 3  \\
2 & 1 & 3
\end{array}\right)$ move two elements.  $\left(\begin{array}{ccc}
1 & 2 & 3  \\
3 & 1 & 2
\end{array}\right)$ and $\left(\begin{array}{ccc}
1 & 2 & 3  \\
2 & 1 & 3
\end{array}\right)$ move three elements.  Just look at how many are left where they started.  With Arturo's answer, you should be on your way.
A: Imagine you have 36 volumes of an encyclopedia in front of you. A permutation $\sigma\in S_{36}$ is a set of instructions of how to swap positions of the volumes of that encyclopedia. For instance, the permutation $\sigma=(1,2,3,4)$  (this is cycle notation: it says element 1 is sent to element 2, element 2 is sent to element 3, 3 to 4, and 4 to 1; if you are used to two-line or abbreviated two-line notation, then this is the permutation that has
$$\left(\begin{array}{cccccccc}
1 & 2 & 3 & 4 & 5 & 6 & \cdots & 36\\
2 & 3 & 4 & 1 & 5 & 6 & \cdots & 36
\end{array}\right).$$
as its two-line description) 
tells you to take volumes $1$ through $4$, and move volume $1$ to the second position, volume $2$ to the third position, volume $3$ to the fourth position, and volume $4$ to the first position. Even though $\sigma$ is a permutation of all 36 volumes, in fact you only need to move four of the volumes to perform the permutation $\sigma$. So we say that $\sigma$ fixes 32 volumes (volumes 5 through 36 are left untouched) and moves 4 volumes (volumes 1 through 4). 
For each permutation $\sigma\in S_n$, which acts on the set $[n]=\{1,2,\ldots,n\}$, you can ask for the set $\mathrm{Fix}(\sigma) = \{ a\in[n]\mid \sigma(a)=a\}$, the set of elements fixed by $\sigma$.
What are the elements "moved" by $\sigma$? They are the elements of $[n]-\mathrm{Fix}(\sigma)$, the complement of $\mathrm{Fix}(\sigma)$.
A $\sigma\in S_{36}$ will lie in $H$ if and only if $[36]-\mathrm{Fix}(\sigma)$ has at most $4$ elements; that is, if and only if $\sigma$ fixes at least 32 of the elements of the numbers in $[36]=\{1,2,3,\ldots,36\}$.
The question is whether $H$ is a subgroup. It is easy to check that it is nonempty and closed under inverses (though you should do that). The trick is going to be figuring out if it is closed under products: if you have $\sigma$ and $\tau$, and each of them fix at least $32$ elements (though possibly not the same 32 elements), can $\sigma\tau$ fix fewer than 32 elements?
