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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.2012-11-18

2 Answers 2

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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 $G$?2012-11-18
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    (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.2012-11-18
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    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 $(2,5)$.2012-11-18
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    You don't really include single element cycles in a cycle decomposition.2012-11-18
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    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 $S_5$?2012-11-18
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    @sizz, Not just to $\sigma$ but to $\sigma^2, \sigma^3,$ and $\sigma^4$.2012-11-18
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    @Brian re:single element cycles: It's a habit I picked up so I/my readers don't forget whether "(123)" is an element of S_5 or S_6, etc. It also avoids the confusion that might arise because I don't know whether sizz uses $\mathbf{1}$ or $e$ or $\mathrm{id}$ for identity elements. But you're right to point out it's non-standard and wasteful when there's no chance for confusion. **sizz**, you also need to multiply by the identity, which is also part of the subgroup generated by anything, and which coincidentally is $\sigma^5$. Also, just be careful: $(6,3,2)$ isn't part of $S_5$2012-11-18
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    Actually, I could stand to be a little clearer. Since in every group (including $S_5$), elements have inverses, if you take *any* single element $g$ in the group, and you want to get an arbitrary element $h$, by multiplication, it's like you want to find the $k$ such that $gk=h$. But the answer to that is just $k=g^{-1}h$. So yes, you *can* get every element of $S_5$ just by multiplying things by $\sigma$. However, the point here is that you're not interested in getting all the elements of $S_5$ element by element, you want *cosets* to generate $S_5$...2012-11-18
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    If $G=\langle\sigma\rangle$, which has five elements, then the *cosets* you can get from any element of $G$ would be the same as $G$: $(1,3,4,2,5)G=G$, etc. And the coset you'd get from $(2,3)$ would be the same as the coset you'd get from $(153)(24)$ since $\left((153)(24)\right)(1,3,4,2,5)=(2,3)$. As you go through the elements of $S_5$, you generate each 5-element coset 5 separate times; there are 24 of them. Cosets *never* overlap: http://math.stackexchange.com/questions/99421/ and have the same size (finite grps): Thm 4.9 in http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/coset.pdf2012-11-18
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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.