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  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 $S_6$

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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 $\mu$?

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There are no elements in $\mu$. It's not a set. – Ink Nov 18 '12 at 7:35
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}.

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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
You don't really include single element cycles in a cycle decomposition. – Ink 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.

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