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List the left cosets $(gH)$ and right cosets $(Hg)$ for $H = \langle (123) \rangle$, where $H \le G$ and $G = S_3$.

My work so far:

$G = S_3 = \langle (12) (13) \rangle = \{ e, (12), (13), (23), (123), (132) \} $

$H = \langle (123) \rangle$

I have trouble listing out all the elements with $H$, as I'm rather iffy with generators and permutations put together. I believe this would help me tremendously on finding the cosets.

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    $\begingroup$ You are new to cosets, so how about this: (i) list out all of the elements of $H$, (ii) for all six elements of $S_3$ compute the corresponding coset of $H$, (iii) identify all redundancies in the resulting handful of six cosets (numerically, there should only be two distinct cosets). $\endgroup$
    – anon
    Feb 19 '14 at 5:50
  • $\begingroup$ That's what I'm having trouble with precisely right now: listing all the elements of $H$. My initial guess is $H = \{e, (123), (132)\}$ ... $\endgroup$
    – Cookie
    Feb 19 '14 at 5:54
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    $\begingroup$ That's correct. First you start with $e$ and $(123)$. You square $(123)$ and get $(132)$. Apply $(123)$ again and you are back at $e$. So you are done - those are the elements of $H$. Now, about those cosets ... you already have the left cosets $eH$, $(123)H$ and $(132)H$. (Do you understand what they are and why you have them?) Compute the other "three" (note - they will all be the same). $\endgroup$
    – anon
    Feb 19 '14 at 5:55
  • $\begingroup$ Just to be sure that I firmly understand this: I also have another part of the exercise where $G = S_3$ still but now I get a different $H$, that is, $H = \langle (12) \rangle$. When I first multiplied $e$ by $(12)$, of course I get $(12)$. Now if I do $(12)(12)$, then I got back to $e$. So would it be $H = \langle (12) \rangle = \{e, (12) \}$? $\endgroup$
    – Cookie
    Feb 19 '14 at 6:02
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    $\begingroup$ Going back to using $H = \langle (123) \rangle$ and $G = S_3$, my left cosets are $(12)H = \{(12), (23), (13) \}$ and $(123)H = \{(123),(132),e\}$. My right cosets are $H(12) = \{(12),(13),(23) \}$ and $H(123) = \{(123), (132), e \}$. I just realized that all the left and right cosets are equal to each other (i.e. $(12)H = H(12)$ and $(123)H = H(123)$, so $H$ is also a normal subgroup! $\endgroup$
    – Cookie
    Feb 19 '14 at 6:21
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Given $G = S_3 = \{e,(12),(13),(23),(123),(132)\}$ and $H = \langle 123 \rangle = \{e,(123),(132)\}$, we have $$eH=\{e,(123),(132)\} \text{ and } He = \{e,(123),(132)\}$$ $$(12)H=\{(12),(23),(13)\} \text{ and } H(12) = \{(12),(13),(23)\}$$ $$(123)H=\{(123),(132),e\} \text{ and } H(123) = \{(123),(132),e\}$$ All cosets of $H$ have been listed. We also see that $eH=He$, $(12)H=H(12)$. and $(123)H=H(123)$. In this case, $H$ is normal subgroup in $G$.

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