# I Why only normal subgroup is used to obtain group quotient

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1. Mar 5, 2016

### SVN

Hello!

As far as I know any subgroup can, in principle, be used to divide group into bundle of cosets. Then any group element belongs to one of the cosets (or to the subgroup itself). And such division still is not qualified as a quotient.

Yes, the bundle of cosets in this case will be different for actions from the right and from the left (although, their number will be the same). But why is that so crucial? We have our division without intersections anyway, do we?

Is there any special name for such «one-sided (pseudo)quotients». Are there any uses for them?

2. Mar 5, 2016

### protonsarecool

For a general group G and a general (non normal) subgroup H, the set of left-cosets G/H, respectively the set of right-cosets H\G won't be groups, you need a normal subgroup (i.e. G/H = H\G) for that.

3. Mar 5, 2016

### SVN

Not quite sure I understand how cosets can be groups themselves. They lack identity element, since it is already included in the generating subgroup (normal or non-normal).

4. Mar 5, 2016

### protonsarecool

You want the set of cosets to be a group (ie. the quotient group).
Say, you have a group $G$ and a subgroup $H$. So if we want to define a product on $G/H$ (where the elements are now left-cosets), we do it like $(aH)\cdot(bH) = (ab)H$. However, this will only make sense iff the left-cosets $aH$ are the same as the right-cosets $Ha$, or $aHa^{-1}=H$. Note however that for a non normal H we have $aHa^{-1} \neq H$. This means in particular that $(aH)\cdot(a^{-1}H)\neq (a a^{-1})H = H$. For a normal H this does work, and the rest of the group axioms are satisfied by $G/H$ with the defined product as well. Only then can we call $G/H$ a quotient group, otherwise its just a set of left-cosets.

5. Mar 6, 2016

### SVN

Thanks a lot for such a detailed explanation. It answers my question fully.

6. Mar 10, 2016

### micromass

Staff Emeritus
In the case of nonnormal $H$, the quotient still has a nice structure of a homogeneous space. That is, there is the obvious group action $G$ on $G/H$ by putting $g\cdot kH = gkH$. This is an important action not only in group theory, but also in geometry since a lot of nice geometries arise as homogeneous spaces.