# Two questions about cycles (algebra)

1. Dec 25, 2012

### Artusartos

I have two questions:

1) For the example on the second page, I don't understand why they say $$\alpha\gamma\alpha^{-1} = (\alpha1 \alpha3)(\alpha2 \alpha4 \alpha7)(\alpha5)(\alpha6)$$ instead of $$\alpha\gamma\alpha^{-1} = (\alpha1\alpha^{-1} \alpha3\alpha^{-1})(\alpha2\alpha^{-1} \alpha4\alpha^{-1} \alpha7\alpha^{-1})(\alpha5\alpha^{-1})(\alpha6\alpha^{-1})$$.

2) For the tables at the top of the 2nd page, I don't know how they computed those numbers...

#### Attached Files:

File size:
13.8 KB
Views:
112
• ###### 20121225_114339.jpg
File size:
32.2 KB
Views:
121
Last edited: Dec 25, 2012
2. Dec 25, 2012

### Stephen Tashi

They say what is consistent with what the theorem says. The theorem says to "apply $\alpha$" to the symbols in the cycles.

If $\alpha,\ p,\ q$ are cycles, It is true that $\alpha (\ p \ q) \ \alpha^{-1} =( \alpha \ p \ \alpha^{-1})(\alpha \ q \ \alpha^{-1})$ but this is not the content of the theorem. A cycle is not the same as the product of the individual symbols in the cycle. The cycle (1,2,3) is not equal to (1)(2)(3).

For example, In the permutation group $S_4$, there are 8 different elements of the group that are cycles of length 3. The example (1,2,3) in the table illustrates one of them.
(There are 24 = (4)(3)(2) different permutations that can be formed by taking 3 distinct numbers from the set of numbers {1,2,3,4}. However, each permutation such as (1,2,3) is one of 3 representations of the same cycle. (1,2,3) = (2,3,1) = (3,1,2) So there are 8 = 24/3 distinct cycles of length 3 )

3. Dec 26, 2012

### algebrat

In a nutshell, notice that alpha gamma alpha inverse takes alpha of 1 to alpha of 3. ;)

4. Dec 26, 2012

Thank you