-412.7.3 decide if cosets of H are the same

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Discussion Overview

The discussion revolves around determining whether specific cosets of the subgroup \( H = \{0; \pm 3; \pm 6; \pm 9; \cdots\} \) are the same. Participants explore the conditions under which two cosets, represented as \( a + H \) and \( b + H \), can be considered equal or distinct, focusing on the differences between the representatives \( a \) and \( b \).

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants propose that two cosets \( aH \) and \( bH \) are equal if and only if \( (b - a) \) is in \( H \).
  • For the cosets \( 11 + H \) and \( 17 + H \), it is noted that \( 17 - 11 = 6 \), which is a multiple of 3, suggesting these cosets are the same.
  • For the cosets \( -1 + H \) and \( 5 + H \), the difference \( 5 - (-1) = 6 \) is also a multiple of 3, indicating these cosets are the same as well.
  • In contrast, for the cosets \( 7 + H \) and \( 23 + H \), the difference \( 23 - 7 = 16 \) is not a multiple of 3, leading to the conclusion that these cosets are not the same.
  • A participant asks for clarification on the condition that the shifts must satisfy for the two cosets to be the same.

Areas of Agreement / Disagreement

Participants generally agree on the method for determining the equality of cosets based on the differences being multiples of 3. However, there is no consensus on the implications of the findings for the specific cases discussed, as the interpretations of the results are not uniformly accepted.

Contextual Notes

The discussion does not resolve the broader implications of the coset equality beyond the specific examples provided, and it relies on the assumption that the subgroup \( H \) is defined as multiples of 3.

karush
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Let $H=\{0;\pm 3;\pm 6;\pm 9;\cdots\}$.
$\textit{ Use }$
$$aH=bH \textit{ or }aH\cap bH=\oslash$$
$\textit{then..}$
$$aH = bH \textit{ iff } (b-a) \textit{ is in } \textit{H}$$
to decide whether or not the following cosets of H are the same.
$\textsf{a. 11 + H and 17 + H}$
$\textsf{b. -1 + H and 5 + H}$
$\textsf{c. 7 + H and 23 + H}$ok not sure what the official method of this would be but for a. 11+6=17 so the coset would would be just a shift over 6 places. the same shift seens to be true for b and c however the beginning numbers would be different

however the book says that c is no
 
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karush said:
Let $H=\{0;\pm 3;\pm 6;\pm 9;\cdots\}$.
$\textit{ Use }$
$$aH=bH \textit{ or }aH\cap bH=\oslash$$
$\textit{then..}$
$$aH = bH \textit{ iff } (b-a) \textit{ is in } \textit{H}$$
to decide whether or not the following cosets of H are the same.
$\textsf{a. 11 + H and 17 + H}$
$\textsf{b. -1 + H and 5 + H}$
$\textsf{c. 7 + H and 23 + H}$ok not sure what the official method of this would be but for a. 11+6=17 so the coset would would be just a shift over 6 places. the same shift seens to be true for b and c however the beginning numbers would be different

however the book says that c is no
$H = \{\ldots -9;-6;-3;\ 0;\ 3;\ 6;\ 9 \ldots\}$

For a.: $11+H = \{\ldots 2;\ 5;\ 8;\ 11;\ 14;\ 17;\ 20;\ \ldots\},\quad 17+H = \{\ldots 8;\ 11;\ 14;\ 17;\ 20;\ 23;\ 25 \ldots\}$
Those two sets are the same. The numbers in the list have just been shifted by two places to get from $11+H$ to $17+H$.

For c.: $7+H = \{\ldots -2;\ 1;\ 4;\ 7;\ 10;\ 13;\ 16;\ 19;\ 22;\ 25;\ 28 \ldots\},\quad 23+H = \{\ldots 14;\ 17;\ 20;\ 23;\ 26 \ldots\}$
Those two sets are not the same. The numbers in them do not coincide at all.

Can you see what condition the shifts must satisfy in order for the two cosets to be the same?
 
Yes: your "method" works. 11+ H and 17+ H are the same because 17- 11= 6 is a multiple of 3; -1+ H and 5+ H are the same because 5- (-1)= 6 is a multiple of 3; and 7+ H and 23+ H are NOT the same because 23- 7= 16 is NOT a multiple of 3.
 

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