# Thread: Cosets and Normal Sub-group: Relation between the two concepts

1. ## Cosets and Normal Sub-group: Relation between the two concepts

We know that for any group, G, a subgroup H is normal-subgroup
IFF every right coset of H is also it's left coset i.e. Ha = aH for all 'a' in G

What if there is a group where every right coset is a left coset but with some other element
i.e. Ha = bH where a=/=b (for all the right cosets of H, obviously there is one exception He = eH)

Basically what I am saying is that there is a one-to-one mapping between set of right cosets and set of left cosets.

I am not sure if such a structure in groups is relevant / important / logical.

What does the above tell the relation between bH and aH? Are they same?

Any thoughts??

2. A related question - Do right cosets and left cosets partition a Group into same mutually exclusive and exhaustive sets?

This is obviously true if H is normal sub-group.

But what if H is not normal is that possible? If yes, what additional property do we have to impose on H?

3. for any subgroup H of a group G these two statements are equivalent:

1) $\displaystyle Ha=aH,$ for all $\displaystyle a \in G.$

2) for any $\displaystyle a \in G,$ there exists $\displaystyle b \in G$ such that $\displaystyle Ha=bH.$

1) implies 2) is trivial. proof of 2) implies 1): so we have $\displaystyle b=ha,$ for some $\displaystyle h \in H.$ then $\displaystyle aH=h^{-1}bH=h^{-1}Ha = Ha.$

i'm not sure i understand your second question!

4. Originally Posted by aman_cc
A related question - Do right cosets and left cosets partition a Group into same mutually exclusive and exhaustive sets?

This is obviously true if H is normal sub-group.

But what if H is not normal is that possible? If yes, what additional property do we have to impose on H?
I don't think I understand either. Maybe this theorem will help you:

Let $\displaystyle H$ be a non-trivial subgroup of G. Let $\displaystyle k = [G:H]$, then:

$\displaystyle G = g_1H \cup g_2H \cup ... \cup g_kH$ where $\displaystyle g_1H, g_2H,...,g_kH$ are the distinct left cosets of H in G. This is also correct for right cosets.

5. Thanks I follow what you say. My questions are wrong. I did not relaize that Ha = Hb is equivalent to Ha = aH. So if every right coset is also a left coset then Ha = aH for all 'a'.

Thanks Defunkt and NonCommAlg

Originally Posted by Defunkt
I don't think I understand either. Maybe this theorem will help you:

Let $\displaystyle H$ be a non-trivial subgroup of G. Let $\displaystyle k = [G:H]$, then:

$\displaystyle G = g_1H \cup g_2H \cup ... \cup g_kH$ where $\displaystyle g_1H, g_2H,...,g_kH$ are the distinct left cosets of H in G. This is also correct for right cosets.