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

[simo1]

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}

 

[Deveno]

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 $\Bbb Z_{10}$?

 

[simo1]

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 $\Bbb Z_{10}$?

my apologies I meant G=S₃

 

[Deveno]

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:

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

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

 

[blue_raver22]

I would explain that G/H represents the quotient group, where the elements of H are the cosets of G. In this case, H is a subgroup of G, meaning that it is a subset of G that also forms a group under the same operation. The elements of G/H are the distinct cosets of H in G, which are represented by the elements α, β, and δ. These elements are not individual elements of G, but rather represent the sets of elements in G that are related by the operation defined by H. Therefore, the elements of G/H are not listed in the same way as the elements of G, but rather as cosets of H.