Difference between revisions of "Coset"
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Latest revision as of 18:08, 15 March 2015
Contents
Definition
Let [ilmath](G,\times)[/ilmath] be a group and [ilmath](H,\times)[/ilmath] a subgroup we denote cosets as follows:
Given any [math]g\in G[/math] the
- Left coset (the left coset of [ilmath]H[/ilmath] in [ilmath]G[/ilmath] with respect to [ilmath]g[/ilmath])
- is denoted [math]gH=\{gh|h\in H\}[/math]
- Right coset (the right coset of [ilmath]H[/ilmath] in [ilmath]G[/ilmath] with respect to [ilmath]g[/ilmath])
- is denoted [math]Hg=\{hg|h\in H\}[/math]
It is quite simply: the set of everything in [ilmath]H[/ilmath] (pre/post) multiplied by [ilmath]g[/ilmath]
Properties
These will be stated for the left coset definition, but the right version is basically the same
Membership
To say [math]x\in gH[/math] is to say [math]\exists y\in H:x=gy[/math] that is:
- [math][x\in gH]\iff[\exists y\in H:x=gy][/math]
Cosets are either disjoint or equal
Given two cosets, [math]g_1H[/math] and [math]g_2H[/math] we have either [math]g_1H=g_2H[/math] or [math]g_1H\cap g_2H=\emptyset[/math]
Suppose they are not disjoint[math]\implies[/math] they are equal
- If they are not disjoint, then take any [math]k\in g_1H\cap g_2H[/math], this means [ilmath]k[/ilmath] is in both [ilmath]g_1H[/ilmath] and [ilmath]g_2H[/ilmath] so:
- [math]k\in g_1H\implies \exists z_1\in H:k=g_1z_1[/math]
- [math]k\in g_2H\implies \exists z_1\in H:k=g_2z_2[/math]
- This means [math]g_1z_1=g_2z_2[/math] allowing us to say:
- [math]g_1=g_2z_2z_1^{-1}[/math] where [math]z_2z_1^{-1}\in H[/math]
- [math]g_2=g_1z_1z_2^{-1}[/math] where [math]z_1z_2^{-1}\in H[/math]
- We now need to show equality of sets
- [math]g_1H\subseteq g_2H[/math], that is [math]x\in g_1H\implies x\in g_2H[/math]
- Take [math]x\in g_1H[/math], this means [math]x=g_1y_1[/math] for some [math]y_1\in H[/math]
- Using [math]g_1=g_2z_2z_1^{-1}[/math] we see [math]x=g_2z_2z_1^{-1}y_1[/math]
- but [math]z_2z_1^{-1}y_1\in H[/math] so let [math]h_1=z_2z_1^{-1}y_1\in H[/math] then
- [math]x=g_2h_1\in g_2H[/math]
- we have shown [math]x\in g_1H\implies x\in g_2H[/math] if they are not disjoint, thus [math]g_1H\subseteq g_2H[/math]
- [math]g_2H\subseteq g_1H[/math], that is [math]x\in g_2H\implies x\in g_1H[/math]
- Take [math]x\in g_2H[/math], this means [math]x=g_2y_2[/math] for some [math]y_2\in H[/math]
- Using [math]g_2=g_1z_1z_2^{-1}[/math] we see [math]x=g_1z_1z_2^{-1}y_2[/math]
- but [math]z_1z_2^{-1}y_2\in H[/math] so let [math]h_2=z_1z_2^{-1}y_2\in H[/math] then
- [math]x=g_1h_2\in g_1H[/math]
- we have shown [math]x\in g_2H\implies x\in g_1H[/math] if they are not disjoint, thus [math]g_2H\subseteq g_1H[/math]
- [math]g_1H\subseteq g_2H[/math], that is [math]x\in g_1H\implies x\in g_2H[/math]
- Combining these we see [math]g_1H\cap g_2H\ne\emptyset\implies g_1H=g_2H[/math]
Suppose they are disjoint[math]\implies[/math] they are not equal
- If they are disjoint then trivially the sets [math]g_1H[/math] and [math]g_2H[/math] are not equal.
So we may conclude [math][g_1H=g_2H]\iff[g_1H\cap g_2H\ne\emptyset][/math]
