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Mathematical Methods for Physicists: A concise introduction - Site Map

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ELEMENTS OF GROUP THEORY<br />

Obviously these two operators <strong>for</strong>m a group that is isomorphic to G 2 . These two<br />

groups (<strong>for</strong>med by the elements E 0 , R, and the elements ^O E 0 and ^O R , respectively)<br />

are the two representations of the abstract group G 2 . These two simple examples<br />

cannot illustrate the value and beauty of the group theoretical method, but they<br />

do serve to illustrate the key concept of isomorphism.<br />

Group of permutations and Cayley's theorem<br />

In Example 12.6 we examined brie¯y the group of permutations of three objects.<br />

We now come back to the general case of n objects (1; 2; ...; n) placed in n boxes<br />

(or places) labeled 1 , 2 ; ...; n . This group, denoted by S n , is called the symmetric<br />

group on n objects. It is of order n! How do we know? The ®rst object may<br />

be put in any of n boxes, and the second object may then be put in any of n 1<br />

boxes, and so <strong>for</strong>th: n…n 1†…n 2†3 2 1 ˆ n!:<br />

We now de®ne, following common practice, a permutation symbol P<br />

P ˆ 1 2 3 n<br />

!<br />

; …12:1†<br />

1 2 3 n<br />

which shifts the object in box 1 to box 1 , the object in box 2 to box 2 , and so<br />

<strong>for</strong>th, where 1 2 n is some arrangement of the numbers 1; 2; 3; ...; n. The old<br />

notation in Example 12.6 can now be written as<br />

<br />

‰2 31Šˆ 1 2 3 <br />

:<br />

2 3 1<br />

For n objects there are n! permutations or arrangements, each of which may be<br />

written in the <strong>for</strong>m (12.1). Taking a speci®c example of three objects, we have<br />

<br />

P 1 ˆ 1 2 3 <br />

; P 2 ˆ 1 2 3 <br />

; P 3 ˆ 1 2 3 <br />

;<br />

1 2 3<br />

2 3 1<br />

1 3 2<br />

<br />

P 4 ˆ 1 2 3 <br />

; P 5 ˆ 1 2 3<br />

2 1 3<br />

3 2 1<br />

<br />

<br />

; P 6 ˆ 1 2 3 <br />

:<br />

3 1 2<br />

For the product of two permutations P i P j …i; j ˆ 1; 2; ...; 6†, we ®rst per<strong>for</strong>m the<br />

one on the right, P j , and then the one on the left, P i . Thus<br />

<br />

P 3 P 6 ˆ 1 2 3 <br />

<br />

1 2 3<br />

ˆ 1 2 3 <br />

ˆ P 4 :<br />

1 3 2 3 1 2 2 1 3<br />

To the reader who has diculty seeing this result, let us explain. Consider the ®rst<br />

column. We ®rst per<strong>for</strong>m P 6 , so that 1 is replaced by 3, we then per<strong>for</strong>m P 3 and 3<br />

438

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