# Left cosets partition the parent group

Given a group $$G$$ and a sub­group $$H$$, the left cosets of $$H$$ in $$G$$ par­ti­tion $$G$$, in the sense that ev­ery el­e­ment of $$g$$ is in pre­cisely one coset.

# Proof

Firstly, ev­ery el­e­ment is in a coset: since $$g \in gH$$ for any $$g$$. So we must show that no el­e­ment is in more than one coset.

Sup­pose $$c$$ is in both $$aH$$ and $$bH$$. Then we claim that $$aH = cH = bH$$, so in fact the two cosets $$aH$$ and $$bH$$ were the same. In­deed, $$c \in aH$$, so there is $$k \in H$$ such that $$c = ak$$. There­fore $$cH = \{ ch : h \in H \} = \{ akh : h \in H \}$$.

Ex­er­cise: $$\{ akh : h \in H \} = \{ ar : r \in H \}$$.

Sup­pose $$akh$$ is in the left-hand side. Then it is in the right-hand side im­me­di­ately: let­ting $$r=kh$$.

Con­versely, sup­pose $$ar$$ is in the right-hand side. Then we may write $$r = k k^{-1} r$$, so $$a k k^{-1} r$$ is in the right-hand side; but then $$k^{-1} r$$ is in $$H$$ so this is ex­actly an ob­ject which lies in the left-hand side. <div><div>

But that is just $$aH$$.

By re­peat­ing the rea­son­ing with $$a$$ and $$b$$ in­ter­changed, we have $$cH = bH$$; this com­pletes the proof.

# Why is this in­ter­est­ing?

The fact that the left cosets par­ti­tion the group means that we can, in some sense, “com­press” the group $$G$$ with re­spect to $$H$$. If we are only in­ter­ested in $$G$$ “up to” $$H$$, we can deal with the par­ti­tion rather than the in­di­vi­d­ual el­e­ments, throw­ing away the in­for­ma­tion we’re not in­ter­ested in.

This con­cept is most im­por­tantly used in defin­ing the quo­tient group. To do this, the sub­group must be nor­mal (proof). In this case, the col­lec­tion of cosets it­self in­her­its a group struc­ture from the par­ent group $$G$$, and the struc­ture of the quo­tient group can of­ten tell us a lot about the par­ent group.

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