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I know $S_4=\{(),(3,4),(2,3),(2,3,4),(2,4,3),(2,4),(1,2),(1,2)(3,4),(1,2,3),(1,2,3,4),(1,2,4,3),(1,2,4),(1,3,2),(1,3,4,2),(1,3),(1,3,4),(1,3)(2,4),(1,3,2,4),(1,4,3,2),(1,4,2),(1,4,3),(1,4),(1,4,2,3),(1,4)(2,3) \}$ and I know there are 8 cosets to be found (I could be wrong) since $24 \div 3=8$ but I have trouble finding out how solve this. Can someone please help? I don't understand how I should multiply each $H$ with the corresponding S₄.

isomorphismes
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isra abuhasna
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  • Choose a permutation not in $H$ and then multiply it on the right of every element in $H$ or on the left (not both at the same time). – graydad Sep 21 '14 at 23:21
  • whoops, my algebra is rusty. Thought cosets still had to be subgroups. – graydad Sep 21 '14 at 23:26

1 Answers1

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For example: $$ (34)H = \{(34)(1),(34)(123),(34)(132)\} = \{(34),(124),(1432)\} $$ is the left coset of $H$ associated with $(3,4)$. We "multiply $H$" by (in this case, left-) multiplying each element in $H$ by the relevant element of $S_4$.

Note that some cosets end up being the same. For example, $$ [(34)(123)]H = (124)H = \{(124),(1432),(34)\} = (34)H $$

You are indeed correct about the number of (distinct) cosets; this is a consequence of Lagrange's theorem.

Ben Grossmann
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    Does this answer your question? If not, could you clarify on where this falls short? Do you need elaboration on how to multiply elements? – Ben Grossmann Sep 21 '14 at 23:30
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    This helps a lot . Multiplying the elements kinda throws me off so can you check if this right ? (23)H gives {(23),(13),(12)}? – isra abuhasna Sep 21 '14 at 23:39
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    That's right. To help with multiplying: try decomposing into cycles straight off. So, for example, to multiply $(23)(132)$: $$ \begin{cases} 1 \overset{(132)}{\to} 3 \overset{(23)}{\to} 2\ 2 \to 1 \to 1 \end{cases} \implies (12) \text{ is one of the cycles}\ 3 \to 2 \to 3 \implies (3) \text{ is one of the cycles} $$ – Ben Grossmann Sep 21 '14 at 23:43