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A permutation of a permutation, is the product of the two permutations?

Right. The multiplication is not commutative, i.e in general

*ab*≠

*ba*. For example $$\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$$ but $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}.$$

What happens if I keep permuting using the same permutation?

If you keep permuting with same permutation, you will eventually return to the initial configuration. For example $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix} =\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix} =\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}.$$ The identity permutation, usually denoted by (1), is the permutation that changes nothing. We know that, say, there are

*n*objects, then applying the same permutation for

*n!*times will always end up to be the identity permuation, i.e. $$\tau^{n!}=(1)$$

Can I say τ

^{n!-1}=1/τ?

Not quite. But you can write τ

^{n!-1}=τ

^{-1}, meaning that it is the multiplicative inverse of τ in the permutation multiplication.

Muliplication with an identity element and inverses. It is a group!?

Right. It is a group.

For every element α of a group

*G*, we define a function

*A*

_{α}(g)=αg. It is not hard to see that

*A*

_{α}is a bijection and so it is a permutation. Indeed we have

*A*

_{α}

*A*

_{β}=

*A*

_{αβ}. Hence if we identify α with

*A*

_{α}, we see that each element of a group is indeed a permutation of the group!