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I'm having a hard-time understanding cosets when the set in question is uncountable. Here's a simple example that leads me to a contradiction.

Let $G = \mathbb{R}\backslash\{0\}$. Then $(G,\cdot)$ is a group where $\cdot()$ is simple multiplication. Furthermore, $H = \{h \in G: -1\le h\le 1\}$ is a subgroup. I'm trying to figure out how $H$ partitions $G$ into its right cosets.

By definition (I'm using the text Elements of Abstract and Linear Algebra), given $a\in G$, then $Ha = \{b\in G: ab^{-1}\in H\} = \{ha:h\in H\}$. But I'm having two problems:

  1. Somehow the two definitions of $Ha$ seem to contradict each other: Say $a = 2$, then the left-hand side means $Ha = \{b\in G: ab^{-1}\in H\} = \{b \in \mathbb{R}: -1 \le ab^{-1} \le 1\}$, which when $a = 2$ implies that $Ha$ is the set $(-\infty,-2]\cup[\infty,2)$. But if I use the right-hand definition I get $Ha = \{ha:-1\le h\le 1\}$ which when $a = 2$ implies that $Ha$ is the set $[-2,2]$. So the two definitions contradict each other.

  2. Moreover, taking $[-2,2]$ as a right coset for $G$, I still fail to see how $H$ would partition $G$, because if I had chosen $a = 3$ for example, then I would have gotten the coset $[-3,3]$ which overlaps with the coset I got with $a = 2$, without being equal to it. So this contradicts that right cosets don't overlap.

What am I misunderstanding? Thanks,

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up vote 5 down vote accepted

$H$ is not a subgroup of $G$. For example, $H$ contains $1/2$ but does not contain $(1/2)^{-1}=2$. So naturally constructions that require $H$ to be a subgroup can't be expected to work.

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Thanks, I knew I was overlooking something simple. I feel like an idiot now. – mahin Dec 9 '11 at 15:47

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