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I have an exercise where I am supposed to find the left and right cosets of H = {(1), (12), (34), (12) ○ (34)} in S4. But how do I generate the cosets? As I have understood it you are supposed to pick a number that is not in the set H and multiply it with every number in H. But this does not exactly give the right answer. I would really appreciate it if someone gave an easy to understand explanation of how to generate the left and right cosets.

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    What is Ž above in your subgroup?2012-05-26
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    You don't pick a "number", you pick an element $g$ of your group $G$ ($G = S_4$ in your example). The left coset $gH$ consists of the elements $gh$ as $h$ ranges over all elements of $H$. If you are having trouble with this, you should show us what specific calculations you have done and we can help pinpoint where you might have went astray.2012-05-26
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    The `Ž` was supposed to be an `○`. The calculations I have done so far is selecting `(13)` as an element and multiplying it with `H`. Then I get `(13)` first(obviously), but if I multiply `(13)` with `(12)` things doesn't exacyly become the answer I am looking for. So I would like to know how you should do all the calculations correct, after selecting an element.2012-05-26
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    What permutation do you get when you compute $(13)(12)$? The set $(13) H = \{ (13), (13)(12), (13)(34), (13)(12)(34) \}$ *is* one of the six left cosets of $H$ in $S_4$, regardless of what predetermined answer you are looking for. My guess is that you are either multiplying permutations incorrectly or your answer has misidentified left and right cosets.2012-05-26
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    I think I have some problems understanding how the multiplications are done... If you take `13` and multiply it with `12` and get `156`, then that does not exactly seem right. How do you do the multiplications correctly? And thanks for all the help everyone! It's really appreciated!2012-05-27

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Assume your group is $G$ and the subgroup is $H$. By definition $gH$={$gh$|$h\in H$} is a left coset of $H$ respect to $g$, in $G$ and $Hg$={$hg$|$h\in H$} is a right coset of $H$ respect to $g$, in $G$. Here your group is $S_4$={$(),(3,4),(2,3),(2,3,4),(2,4,3),(2,4), (1,2),(1,2)(3,4),(1,2,3),(1,2,3,4),(1,2,4,3),(1,2,4),(1,3,2),(1,3,4,2),(1,3), (1,3,4),(1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3)$ }, and the $H$ is as you pointed. According to Group theory, the number of right cosets of a subgroup in its group called index is $\frac{|G|}{|H|}$. $|S_4|=4!$ and $|H|=|\langle(1,2),(3,4)\rangle|=4$ so you have atlast $\frac{4!}{4}=6$ cosets right or left for the subgroup. Here there is no matter what $g$ is taken in group $G$. For example, if you take $(1,2,4,3)$ in group, then $(1,2,4,3)H$={$(1,2,4,3)(),(1,2,4,3)(1,2),(1,2,4,3)(3,4),(1,2,4,3)(1,2)(3,4)$}. Hope to help.

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    Very thorough, listing all the elements! +12013-03-05
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    $|H|=4$ because it has four elements, correct?2015-03-05
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    @grayQuant: Yes that's right! :-)2015-03-06
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Take for example $\,\pi:= (123)\,$ :$$\pi(1)=\pi\,,\,\pi(12)=(13)\,,\,\pi(34)=(1234)\,,\,\pi(12)(34)=(134)\Longrightarrow \pi H=\{(123),\,(13),\,(1234),\,(134)\}$$

Now try to find $\,H\pi\,$, and check whether you can find examples of $\,\pi\sigma^{-1}\in H\,$ , since then

$\,\pi H=\sigma H\,$ and you can save quite some time.

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Find all distincr right and left costs of the subgroup H={0,2,4} of the group (z6,+6) and obtain the right coset decomposition of z6.