Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. Let $\sigma = (1, 2, 5, 4)(2,3)$ in $S_5$. Find the index of $<\sigma>$ in $S_5$

  2. Let $\mu = (1,2,4,5)(3,6)$ in $S_6$. Find the index of $<\mu>$ in $S_6$

enter image description here

Here is what I don't understand.

1) This seems like an advanced method of counting. They found the number of elements in the group - the order is 5. Then they count the permutations of the big group, which is 5!. Now what I don't understand is dividing them out? I am a little lost as to why they are doing that? Could someone show me just what one coset even looks like here?

2) Same question as (1), I like to see one coset and this is just a refresher for me because I am confusing the order of a group. They say the order of the subgroup is 4 because they are disjoint because it takes 4 mappings for $\mu$ to map back to the identity, but I thought the order of a group means the number of elements. So if I were to count, isn't there still $6$ elements in $\mu$?

share|improve this question
1  
There are no elements in $\mu$. It's not a set. –  Vectk Nov 18 '12 at 7:35

2 Answers 2

up vote 0 down vote accepted

1) There are important theorems that say cosets are disjoint and have the same size. Therefore, once you believe (1,2,3,5,4) generates a subgroup of order 5, the index must be 5!/5 since the cosets partition the set of group elements into 5s.

The easiest coset to exhibit is the subgroup (call it G) itself: G={(1,2,3,5,4),(1,2,3,5,4)2,(1,2,3,5,4)3,(1,2,3,5,4)4,(1,2,3,5,4)5}={(1,2,3,5,4),(1,3,4,2,5),(1,5,2,4,3),(1,4,5,3,2),(1)(2)(3)(4)(5)}. Another left coset is (2,3)G={(2,3)(1,2,3,5,4),(2,3)(1,3,4,2,5),(2,3)(1,5,2,4,3),(2,3)(1,4,5,3,2),(2,3)(1)(2)(3)(4)(5)}={(1,3,5,4)(2),(1,2,5)(3,4),(1,5,3)(2,4),(1,4,5,2)(3),(1)(2,3)(4)(5)}

2) The order of a group is the number of elements. The things being permuted don't really relate directly. The subgroup generated by μ has four elements: {(1,2,4,5)(3,6),(1,4)(2,5)(3)(6),(1,5,4,2)(3,6),(1)(2)(3)(4)(5)(6)}. It just so happens that the four elements of that group are themselves permutations of the set (not a group) {1,2,3,4,5,6}.

share|improve this answer
    
OKay I am very lost by example. How did you pick (2,3) from $G$? –  Hawk Nov 18 '12 at 7:14
    
(2,3) is not in G. (2,3) is in $S_5$ (and I just picked it randomly), and as such, it gives rise to a left coset of G. If I had picked something in G (say, (1,3,4,2,5)) then the left coset (1,3,4,2,5) G would just be G again. It wouldn't be a new coset. –  Mark S. Nov 18 '12 at 7:17
    
In response to a deleted comment, $\left((1,2,4,5)(3,6)\right)^2$ sends $3$ to $6$ and then back to $3$, so it should have $(3)$. Similarly for $(6)$. However, it sends $1$ to $2$ and then to $4$, and $4$ to $5$ and then to $1$, so it should have $(1,4)$ as part of its disjoint cycle decomposition. Similarly, it sends $2$ to $4$ and then to $5$, and $5$ to $1$ and then to $2$, so it should have $(2,5)$. –  Mark S. Nov 18 '12 at 7:42
1  
You don't really include single element cycles in a cycle decomposition. –  Vectk Nov 18 '12 at 8:14
    
So basically the answer is saying I can take 24 elements in $S_5$ like $(3,1)$, $(6,3,2)$ and multiply it to $\sigma$ and eventually that will exhaust $S_5$? –  Hawk Nov 18 '12 at 8:16

If $H \subseteq G$ is a subgroup, its index $[G: H$] is the number of left cosets of $H$ in $G$. Recall that a left coset is a set $gH = \{gh : h \in H\}$ for some $g \in G$. The left cosets partition $G$ and $|gH| = |H|$ for any $g$, so if $G$ is finite $[G: H] = |G|/|H|$.

As for your second question, it looks like you need to review what the symmetric group is.
You are correct that the order of a group is the number of elements in the group. In this case $\mu$ is an element of $S_6$, and the group generated by it is

$$\{\mu, \mu^2, \mu^3, id\} = \{(1245)(36), (14)(25), (1542)(36), id \}$$

which has four elements.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.