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.


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.