Problem regarding proving a permutation group The question states: 

Show that the set of permutations of three objects form a group. Give the multiplication table for this group.

If we take three distinct objects, the set of the permutations will have cardinality 6. How do we prove it's a group? And how do we make a multiplication table? I'm new to groups so I don't have a firm grasp over the concepts.
 A: Getting some notation should help.
Let the set of three objects be $\{1,2,3\}$. An example of a permutation then is $(12)$. This notation means that object $2$ takes the place of object $1$, and object $1$ takes the place of object $2$. An example of a permutation of three objects is $(123)$, meaning $3$ takes the place of $2$, $2$ takes the place of $1$, and $1$ takes the place of $3$. In general, we work from right to left.
An example of multiplication is $(12)(23)$. To determine what this equals, we see how it acts on each element of the set. What happens when we perform this permutation on object $1$? Working from right to left, we see that the first permutation does not affect object $1$, but the second permutation sends $1$ to $2$. The first permutation sends $2$ to $3$, but the second one doesn't affect $3$. Lastly, the first permutation sends $3$ to $2$, and the second permutation sends $2$ to $1$. So $1$ gets sent to $2$, $2$ gets sent to $3$, and $3$ gets sent to $1$. Hence, we have $(12)(23) = (321)$.
Hopefully these examples are illustrative enough to get you started.
A: A group is a set X together with a map $f : X\times X->X$ (If the (ordered!) pair $(a,b)$
 is mapped to $c$, this is denoted with $a*b=c$, where "*" does not necessariliy mean
 the multiplication), such that the following properties hold :
There is an element $e$, called neutral element with the property $x*e=x$ for all $x$
For every $x$, there is a $y$ (called inverse of $x$) with $x*y=e$
For all $x,y,z$, the equation $x*(y*z)=(x*y)*z$ holds (rule of associativity)
Now, in our case, with "*" it is meant that the permutations act one after another.
For example
p : 1 2 3
 2 3 1

q : 1 2 3
 3 2 1

We have 
p*q : 1 2 3
   2 1 3

The neutral element is the idendity permutation 
id : 1 2 3
  1 2 3

A: The group you are considering is $S_{3}$, the symmetries (permutations) over three elements. The operation is function composition. So you need to show a few things to prove it is a group:
-The operation is associative
-The operation is closed (given two elements $\sigma_{1}, \sigma_{2} \in S_{3}$, is $\sigma_{1} \odot \sigma_{2} \in S_{3}$)
-The operation is invertible (for every $\sigma$, there exists $\sigma^{-1}$ such that $\sigma \sigma^{-1} = \sigma^{-1} \sigma = 1$.
-There is an identity element (this is easy- the identity mapping)
So think of permutations as bijections: $\pi : S \to S$, where $S$ is your set.
For invertibility, just undo the mapping. So consider the table:
$$\begin{bmatrix}1 & 2 & 3 \\ \sigma(1) & \sigma(2) & \sigma(3) \end{bmatrix}$$
Giving us a permutation. If $\sigma$ sends $3 \mapsto \sigma(3)$, then $\sigma^{-1}$ must take $\sigma(3) \mapsto 3$. Can you construct a $\sigma^{-1}$ given $\sigma$?
Once you understand how to prove it is a group, the multiplication table should follow easier.
