1

I read that only cosets (G/H) such that H is a normal subgroup form a factor group. Can anyone tell me an example of a case where cosets do not form a group ?

Gerry Myerson
  • 179,216
Jackson Kailath
  • 256
  • Related (but not actually a duplicate since this question is about an example) http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/ – Tobias Kildetoft Aug 28 '15 at 12:48

1 Answers1

4

Take any group $G$ with a non-normal subgroup. For example $G = S_{3}$ and $H = \langle (1 \ 2) \rangle$. The cosets are $$\{e, (1 \ 2)\}, \ \{e, (1 \ 2)\} \cdot (1 \ 3) = \{(1 \ 3),(1 \ 3 \ 2)\},\ \{e, (1 \ 2)\} \cdot (2 \ 3) = \{(2 \ 3), (1 \ 2 \ 3)\}.$$
Multiplication between cosets is not well defined, because for example $$e \cdot (1 \ 3) =(1 \ 3) \in \{(1 \ 3), (1 \ 3 \ 2)\}$$ but $$(1 \ 2) \cdot (1\ 3\ 2) = (1 \ 2\ 3) \in \{(2 \ 3), (1 \ 2 \ 3)\}.$$

xxxxxxxxx
  • 13,302
  • 1
    Can this be adapted to show that no multiplication can be defined, so that we may be sure the cosets don't form any group? EDIT: answer is clearly no, just set-biject the elements with a cyclic group. – Patrick Stevens Aug 28 '15 at 07:49
  • @Morgan Rodgers can you explain your answer a bit more the part towards the end where you say that multiplication is not well defined.tx in advance. – Jackson Kailath Aug 28 '15 at 11:27