MHB How Do You List Elements of G/H in Z10 When H={α,β,δ}?

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To list the elements of G/H where G is S3 and H is a subgroup generated by a 3-cycle, it is important to note that H consists of the identity and the two 3-cycles. According to Lagrange's Theorem, there will be two cosets: H and Ha, with a transposition like (1 2) serving as a representative for the second coset. The elements of the coset H(1 2) can be calculated to yield the set {(1 2), (1 3), (2 3)}. This illustrates how to effectively list the elements of G/H in this context.
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someone had a post on finite quotient groups. i understood that but how does one list elements of G/H if H is a subgroup of G.
where:
G=Z10 H={α,β,δ}
 
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Re: clarity on groups

$\Bbb Z_{10}$ has no subgroup of 3 elements, since 3 does not divide 10. Did you have in mind some specific subgroup of $\Bbb Z_{10}$?
 
Re: clarity on groups

Deveno said:
$\Bbb Z_{10}$ has no subgroup of 3 elements, since 3 does not divide 10. Did you have in mind some specific subgroup of $\Bbb Z_{10}$?

my apologies I meant G=S3
 
Re: clarity on groups

Well, fortunately, there is only ONE subgroup of $S_3$ of order 3, which is generated by any 3-cycle.

The USUAL notation for such a 3-cycle is:

$(1\ 2\ 3)$ which is shorthand for the mapping:

$1 \to 2$
$2 \to 3$
$3 \to 1$.

The subgroup generated by this is:

$H = \{e, (1\ 2\ 3), (1\ 3\ 2)\}$.

Lagrange's Theorem tells us there will be exactly TWO cosets, $H$ and $Ha$ where $a$ is any permutation not in $H$. The 2-cycle (transposition) $(1\ 2)$ will do, and we find:

$H(1\ 2) = \{(1\ 2), (1\ 2\ 3)(1\ 2), (1\ 3\ 2)(1\ 2)\}$

$= \{(1\ 2), (1\ 3), (2\ 3)\}$
 
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