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

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SUMMARY

This discussion focuses on listing elements of the quotient group G/H where G is the symmetric group S3 and H is the subgroup generated by the 3-cycle (1 2 3). It clarifies that S3 has only one subgroup of order 3, which consists of the identity and the two 3-cycles. By applying Lagrange's Theorem, the discussion concludes that there are exactly two cosets: H and Ha, with Ha being generated by the transposition (1 2), resulting in the set { (1 2), (1 3), (2 3) }.

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  • Familiarity with symmetric groups, particularly S3
  • Knowledge of Lagrange's Theorem and its implications
  • Ability to work with cycle notation in permutations
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Mathematicians, students of abstract algebra, and anyone interested in advanced group theory concepts will benefit from this discussion.

simo1
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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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