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The answer is:

There are 24 permutations. The 12 even permutations are: id , (1 2 3 4) , (1 3 2 4) , (1 4 2 3) , (1 2 3) , (1 2 4) , (1 3 2) , (1 3 4) , (1 4 2) , (1 4 3) , (2 3 4) , (2 4 3).

The 12 odd ones: (1 2) , (1 3) , (1 4) , (2 3) , (2 4) , (3 4) , (1 2 3 4) , (1 2 4 3) , (1 3 2 4) , (1 3 4 2) , (1 4 2 3) , (1 4 3 2).

The question is, why we simply do not write out all the combinations as follows:

(1 2 3 4) , (1 2 4 3) , (1 4 2 3) , ect.?

Also how do we get combinations such as (1 2) , (1 3) , (1 4 3) and so on?

Thanks

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  • $\begingroup$ What do you mean by 'even' and 'odd' in this case? Are you referring to parity? $\endgroup$ – Graham Kemp Apr 3 '14 at 22:40
  • $\begingroup$ Do you know what the cycle notation is? Because you seem to be using it. You might be thinking about a matrix notation $$\begin{pmatrix} 1&2&3&4\\2&1&3&4\end{pmatrix}$$ but that is very clumsy. Note that is $(12)$. $\endgroup$ – Pedro Tamaroff Apr 3 '14 at 22:40
  • $\begingroup$ @GrahamKemp The sign of the permutation. $\endgroup$ – Pedro Tamaroff Apr 3 '14 at 22:40
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    $\begingroup$ $(1\,2\,3\,4)$ is odd, not even. An even permutation is a product of an even number of transpositions, and $(1\,2\,3\,4) = (1\,2)(1\,3)(1\,4)$, which is three transpositions. $\endgroup$ – MJD Apr 3 '14 at 22:43
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    $\begingroup$ $(1\; 2)(3\; 4) \equiv \begin{pmatrix}1&2&3&4\\2&1&4&3\end{pmatrix}; \quad (1\; 2\; 3\; 4) \equiv \begin{pmatrix}1&2&3&4\\2&3&4&1\end{pmatrix}$ $\endgroup$ – Graham Kemp Apr 3 '14 at 23:09
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It is a matter of cycle notation.

$\{1, 2, 3, 4\}$ is the set.  $(1\; 2\; 3\; 4)$ is a permutation of the set; a sequence of transpositions of that set represented by the rotation of elements.  (A transposition is a permutation resulting from an exchange of just two elements.)

This permutation can also be expressed as: $\begin{pmatrix}1 & 2& 3 &4\\ 2 & 3 & 4 & 1\end{pmatrix}$ (Cauchy notation; initial state above, end state below).

It means that the first element is exchanged with the second, then the third, and finally the fourth.  This is also expressed as a product of the individual transpositions: $(1\; 2\; 3\; 4) = (1\; 2)(1\; 3)(1\; 4)$  That is an odd number of transpositions (three), meaning it has an odd parity.

$(1 \; 2)$ is a single transposition; so it's an odd parity. $(1\; 2\; 4)$ is two transpositions; so it's an even parity.

$(1\; 2) \equiv \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4\end{pmatrix}$

$(1\; 2\; 3) = (1\; 2)(1\; 3) \equiv \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4\end{pmatrix}$

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