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

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Discussion Overview

The discussion revolves around how to list the elements of the quotient group G/H, where G is a group and H is a subgroup. The specific groups mentioned are G=Z10 and later G=S3, with H being a subgroup of G.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how to list elements of G/H when H is a subgroup of G, specifically mentioning G=Z10 and H={α,β,δ}.
  • Another participant points out that Z10 has no subgroup of 3 elements, as 3 does not divide 10, and asks if a specific subgroup was intended.
  • A later reply clarifies that the intended group is G=S3, correcting the earlier mention of Z10.
  • It is noted that S3 has only one subgroup of order 3, generated by a 3-cycle, and provides the specific elements of that subgroup.
  • Discussion includes the application of Lagrange's Theorem, indicating that there will be exactly two cosets for the subgroup of S3.

Areas of Agreement / Disagreement

Participants do not reach consensus on the initial subgroup mentioned, as there is confusion regarding the groups involved. The discussion transitions from Z10 to S3, indicating a shift in focus but not resolving the initial query about G/H.

Contextual Notes

The discussion highlights limitations in understanding subgroup structures, particularly regarding the order of elements and the application of group theory concepts like Lagrange's Theorem.

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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