1

Problem: Find a set of representative of the left cosets of $H$ in $S_4$, where $H = \left\{ \sigma \in S _ { 4 } \mid \sigma ( 4 ) = 4 \right\}$.

Solution: In the case of $S _ { 4 }$ we have $\{ \mathrm { id } , ( 14 ) , ( 24 ) , ( 34 ) \}$ as a set of representatives (or transversal).

I have trouble understanding intuitively the concept of cosets. I do understand that a representative is a member of the coset. How $(14)$ is a representative of a coset in $S_4$? What coset is it representing?

Bernard
  • 175,478
NotAbelianGroup
  • 739

1 Answers1

1

$(1\,4)$ is a representative of the left coset… $(1\,4)H$.

Also, two permutations $\rho,\sigma\in S_4$ define the same coset modulo $H$ if and only if $\rho^{-1}\sigma\in H$, i.e. if the composition $\rho^{-1}\sigma$ leaves $4$ fixed. Now you can easily check this condition is not satisfied by any of the pairs $\{\operatorname{Id}, (1\,4), (2\,4), (3\,4)\}$, e.g. $$(1\,4)(2\,4)=(1\,4\,2)$$ Finally, observe $H$ is isomorphic to $S_3$ (it permutes $1,2,3$ between them). So the set $S_4/H$ has cardinal $$\bigl|S_4/H\bigr|=\frac{\bigl|S_4\bigr|}{\bigl|S_3\bigr|}=4,$$ so we've obtained the exact number of cosets representatives.

Bernard
  • 175,478
  • Thank you very much, I was trying to understand how the set of representative was found and you explained it really well. @Bernard how the criteria $\rho , \sigma \in S _ { 4 }$ define the same coset modulo $H$ if and only if $\rho ^ { - 1 } \sigma \in H$ can be extended to group actions? – NotAbelianGroup Dec 26 '18 at 16:43
  • I don't understand very well what situation in group actions your question refers to? Could explain with more details? – Bernard Dec 26 '18 at 16:51
  • Sorry I did a mistake, it wasn't about group actions but quotient groups. Let $G$ a finite group, $H$ a normal subgroup of $G$, we know that $G/H$ (the set of all cosets of H in G) is a group. To expand on your comment, can we say that $g_1$ and $g_2$ define the same coset modulo H if and only if $g_1 ^{-1} g_2 \in H$? – NotAbelianGroup Dec 26 '18 at 17:02
  • Yes absolutely. The only difference is that $G/H$ is a group when $H$ is a normal subgroup, and that left cosets are the same as right cosets. – Bernard Dec 26 '18 at 17:05