User:Peterungar/sandbox

a = (12)(34) b = (13)(2)(4)  While we can think of a single permutation as a rearrangement of our objects, 1,2,3 and 4 in the present case, it is easier to understand permutation groups if we think of a permutation $$P$$ of n objects as a mapping of the set of objects onto itself. We can present it as a table

$$ P = \tbinom{a_1\ a_2 \ \dots \ a_n}{b_1\ b_2\ \dots \ b_n}$$.

If $$ Q $$ is another permutation

$$ Q = \tbinom{b_1\ b_2\ \dots \ b_n}{c_1\ c_2\ \dots \ c_n}\ $$

the composition or product of the permutations is defined as

$$\ PQ = \tbinom{a_1\ a_2\ \dots \ a_n}{c_1\ c_2 \ \dots \ c_n}$$.

Note that the convention for the composition of permutations is the opposite of the convention used when forming the composition f∘g  of two functions. In PQ we apply P first and Q afterwards, but when we form  f∘g  we apply the second function g first.

Example. Let

$$ P = \tbinom{1\ 2\ 3\ 4}{2\ 1\ 4\ 3},\ \ Q = \tbinom{2\ 1\ 4\ 3}{2\ 3\ 4\ 1} = \tbinom{1\ 2\ 3\ 4}{3\ 2\ 1\ 4}$$.

Then $$ PQ = \tbinom{1\ 2\ 3\ 4}{2\ 3\ 4\ 1}\ $$  and   $$\  QP = \tbinom{1\ 2\ 3\ 4}{4\ 1\ 2\ 3}.$$

PQ is also called the product of the group elements P and Q. As we just saw, this kind of multiplication is not usually commutative.