Coset

Definition

Let $(G,\times)$ be a group and $(H,\times)$ a subgroup we denote cosets as follows:

Given any $$g\in G$$ the

• Left coset (the left coset of $H$ in $G$ with respect to $g$)

  is denoted $$gH=\{gh|h\in H\}$$

• Right coset (the right coset of $H$ in $G$ with respect to $g$)

  is denoted $$Hg=\{hg|h\in H\}$$

It is quite simply: the set of everything in $H$ (pre/post) multiplied by $g$

Properties

These will be stated for the left coset definition, but the right version is basically the same

Membership

To say $$x\in gH$$ is to say $$\exists y\in H:x=gy$$ that is:

• $$[x\in gH]\iff[\exists y\in H:x=gy]$$

Cosets are either disjoint or equal

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Given two cosets, $$g_1H$$ and $$g_2H$$ we have either $$g_1H=g_2H$$ or $$g_1H\cap g_2H=\emptyset$$

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Suppose they are not disjoint $$\implies$$ they are equal

If they are not disjoint, then take any $$k\in g_1H\cap g_2H$$ , this means $k$ is in both $g_1H$ and $g_2H$ so:

• $$k\in g_1H\implies \exists z_1\in H:k=g_1z_1$$
• $$k\in g_2H\implies \exists z_1\in H:k=g_2z_2$$

This means $$g_1z_1=g_2z_2$$ allowing us to say:

• $$g_1=g_2z_2z_1^{-1}$$ where $$z_2z_1^{-1}\in H$$
• $$g_2=g_1z_1z_2^{-1}$$ where $$z_1z_2^{-1}\in H$$

We now need to show equality of sets

1. $$g_1H\subseteq g_2H$$ , that is $$x\in g_1H\implies x\in g_2H$$

   Take $$x\in g_1H$$ , this means $$x=g_1y_1$$ for some $$y_1\in H$$

   Using $$g_1=g_2z_2z_1^{-1}$$ we see $$x=g_2z_2z_1^{-1}y_1$$

   but $$z_2z_1^{-1}y_1\in H$$ so let $$h_1=z_2z_1^{-1}y_1\in H$$ then

   $$x=g_2h_1\in g_2H$$

   we have shown $$x\in g_1H\implies x\in g_2H$$ if they are not disjoint, thus $$g_1H\subseteq g_2H$$

2. $$g_2H\subseteq g_1H$$ , that is $$x\in g_2H\implies x\in g_1H$$

   Take $$x\in g_2H$$ , this means $$x=g_2y_2$$ for some $$y_2\in H$$

   Using $$g_2=g_1z_1z_2^{-1}$$ we see $$x=g_1z_1z_2^{-1}y_2$$

   but $$z_1z_2^{-1}y_2\in H$$ so let $$h_2=z_1z_2^{-1}y_2\in H$$ then

   $$x=g_1h_2\in g_1H$$

   we have shown $$x\in g_2H\implies x\in g_1H$$ if they are not disjoint, thus $$g_2H\subseteq g_1H$$

Combining these we see $$g_1H\cap g_2H\ne\emptyset\implies g_1H=g_2H$$

Suppose they are disjoint $$\implies$$ they are not equal

If they are disjoint then trivially the sets $$g_1H$$ and $$g_2H$$ are not equal.

So we may conclude $$[g_1H=g_2H]\iff[g_1H\cap g_2H\ne\emptyset]$$

See also

• Normal subgroup