# Coset problem

• Sep 3rd 2008, 05:48 AM
Coset problem
Let H be a subgroup of G, and b a fixed element of G.

Prove the following statements are equviaent.

1. bH = Hb
2. If $\displaystyle c \in G$ such that $\displaystyle cH=bH$, then $\displaystyle Hc = Hb$
3. If $\displaystyle f \in G$ such that $\displaystyle Hf=Hb$, then $\displaystyle fH=bH$

My Proof.

1 to 2:

Let x be in cH, then there exist an element u in H such that $\displaystyle x=cu$. Since $\displaystyle cH=bH$, there exist an element v in H such that $\displaystyle x=cu=bv$. Since $\displaystyle bH=Hb$, there exist an element w in H such that $\displaystyle x=cu=bv=wb$.

Now, suppose that y is in Hc, then there exist an element d in H such that $\displaystyle y=dc$.

I want to get y to look like Hb, but can't really get it...

Thanks.
• Sep 3rd 2008, 07:10 AM
Laurent

Assume 1 (i.e. $\displaystyle bH=Hb$). Let $\displaystyle c$ be such that $\displaystyle cH=bH$. Hence $\displaystyle cH=bH=Hb$.

Notice that $\displaystyle c\in cH=bH=Hb$, so that there are $\displaystyle h,g\in H$ such that $\displaystyle c=bh=gb$.

Now, if $\displaystyle y\in Hc$, $\displaystyle y=dc$ for some $\displaystyle d\in H$, hence $\displaystyle y=d(gb)=(dg)b\in Hb$ (because $\displaystyle H$ is a subgroup), and $\displaystyle Hb=cH$, so that $\displaystyle y\in cH$. We've proved that $\displaystyle Hc\subset cH$. Let's prove the reverse inclusion. If $\displaystyle y\in cH$, then $\displaystyle y=db$ for some $\displaystyle d\in H$ (because $\displaystyle cH=Hb$), hence $\displaystyle y=db=d(g^{-1}c)=(dg^{-1})c\in Hc$ (because $\displaystyle H$ is a subgroup, again). This proves that $\displaystyle cH\subset Hc$. Finally, $\displaystyle Hc=cH$.

Laurent.
• Sep 4th 2008, 09:10 AM
Thanks for the tips, and I have proved from i to iii similiarly.

Now, I'm trying to prove from ii to i:

Let c be in G such that cH=bH, and we have Hc=Hb.

c is in cH and Hc, so there exist elements x,y in H such that c=bx=yb. Implies that $\displaystyle b=cx^{-1}=y^{-1}c$

Suppose that u is in bH, then there exist h in H such that u=bh, I'm trying to get this to look like (xxx)b, am I still on the right track?
• Sep 5th 2008, 12:37 AM
Laurent
Don't forget that (ii) is "for every c such that ..." and not "there exists c such that...", so you'll probably need to apply (ii) to some well chosen c.
The most obvious choice would be to try applying (ii) to the element $\displaystyle u$ you introduced. You can check that it indeed works, and conclude the proof.
• Sep 6th 2008, 01:18 PM
Here is what I have so far:

Let $\displaystyle u=bh$ then $\displaystyle uH=(bh)H=bH$, then I have $\displaystyle Hu=Hb$

I want to show that $\displaystyle uH = Hu$, which would mean that $\displaystyle bH=Hb$

Claim: $\displaystyle uH \subseteq Hu$

Let $\displaystyle x \in uH$, then there exist an element $\displaystyle a \in H$ such that $\displaystyle x = ua = (bh)a$

Am I on the right track here?

Thanks.
• Sep 6th 2008, 05:14 PM
Laurent
Remember you let $\displaystyle u\in bH$, so you want to prove that $\displaystyle u\in Hb$.

Once you have proved $\displaystyle Hu = Hb$, the trivial identity $\displaystyle u\in Hu$ shows you that $\displaystyle u\in Hb$. Hence $\displaystyle bH\subset Hb$.

Two remarks:
1) There is something we should have mentioned first: if $\displaystyle xH\subset yH$, then $\displaystyle xH=yH$. We already proved this once or twice: we have $\displaystyle x=yh$ for some $\displaystyle h\in H$, hence $\displaystyle y=xh^{-1}$ and $\displaystyle yH=xh^{-1}H\subset xH$. Using this, the proof could have been shortened a bit.

2) More important: I wonder if the statement of your problem is correct. The reason is: we proved that (ii) implies $\displaystyle bH\subset Hb$. There remains to prove that $\displaystyle Hb\subset bH$. However, if you look at the proof of (i) implies (ii), we only made use of the inclusion $\displaystyle bH\subset Hb$. Hence we have proved that:
$\displaystyle bH\subset Hb$ is equivalent to: for every $\displaystyle c\in G,$ $\displaystyle cH=bH\Rightarrow Hc=Hb.$
The same way, there is an equivalence between the inverse inclusion and (iii).
So if your statement were exact, (ii) and (iii) would be equivalent, which means that: $\displaystyle bH\subset Hb$ is equivalent to $\displaystyle Hb\subset bH$, or $\displaystyle bHb^{-1}\subset H$ is equivalent to $\displaystyle bHb^{-1}=H$.
If the subgroup $\displaystyle H$ is finite, this is OK ($\displaystyle bH$ and $\displaystyle Hb$ have same cardinality). Otherwise I feel like this is not always true but I can't find a counter-example. I'm going to ask the question in a new thread.
• Sep 7th 2008, 02:03 AM
Laurent
As a conclusion:

The answer (by NonCommAlg) to this thread proves that it is possible to have $\displaystyle bH\subset Hb$ and not $\displaystyle Hb\subset bH$.

For that reason, (ii) and (iii) can't be equivalent.

A possible correct statement would be: For any $\displaystyle b\in G$, the following properties are equivalent:
(i') $\displaystyle bH\subset Hb$;
(ii) For any $\displaystyle c\in G$, if $\displaystyle cH=bH$, then $\displaystyle Hc=Hb$.

And the same with reverse inclusion and (iii).

I spent a bunch of time yesterday trying to prove (i) from (ii) before I realized this was possibly false... Where did you get this exercise from? (or did you forget to mention that $\displaystyle H$ is supposed to be finite?)
• Sep 7th 2008, 12:39 PM