Proof involving group and subgroup

[ArcanaNoir]

Homework Statement

If H is a subgroup of a group G, and a, b $\in$ G, then the following four conditions are equivalent:
i) $ab^{-1} \in H$
ii) $a=hb$ for some $h \in H$
iii) $a \in Hb$
iv) $Ha=Hb$

Homework Equations

cancellation law seems handy, and existence of an inverse, associativity, and other basic things.

The Attempt at a Solution

$ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h \Rightarrow (ab^{-1})b=hb \Rightarrow a(bb^{-1})=ae=a=hb \Rightarrow a \in Hb \Rightarrow ...?$

 

[micromass]

Ok, so you already proved (1)=>(2) and (2)=>(3).

So assume that (3) holds. So a is in bH.

You need to prove that aH=bH. So take an element in aH and prove that it is in bH, and take an element in bH and prove that it is in aH. Write these things out.

 

[ArcanaNoir]

does H=Hb?

 

[micromass]

does H=Hb?

Only if b is in H.

You need to prove that aH is in bH.
So take an element ah in aH, you must prove that it is in bH. So you must write it in the form bh'.

 

[ArcanaNoir]

I'm stuck. I tried doing something like a=hb but for ha=b but since h=ab^-1 that didn't help, so I considered $h^{-1}=ba^{-1}$ so $ba^{-1}=h^{-1}$ and $h=ab^{-1} \Rightarrow$
But I'm not getting anywhere. What should come directly after $a \in Hb \Rightarrow$ ?

 

[micromass]

Take ah in aH. You must show that it is in bH. So you must find h' in H so that ah=bh'.
The logical thing to do is to take $h^\prime=b^{-1}ah$. But why is this in H?

 

[ArcanaNoir]

is it a problem that I'm looking for Ha, not aH?

 

[micromass]

is it a problem that I'm looking for Ha, not aH?

Oh, I'm sorry. I didn't notice that. It doesn't change much.

You need to find h' such that ha=h'b. So pick $h^\prime=hab^{-1}$. but why is this in H?

 

[ArcanaNoir]

okay, h is in H already, so hh is in H, and $hh=hab^{-1}=h' \in H$

 

[ArcanaNoir]

so $h' \in H \Rightarrow h'b \in Hb \Rightarrow ah'b \in Hb \Rightarrow ah' \in H \Rightarrow$
$h^{-1}ah' \in H \Rightarrow ba^{-1}ah' \in H \Rightarrow bh' \in H$

 

[micromass]

hh=hab^{-1}

Why?? This isn't true...

 

[ArcanaNoir]

Why?? This isn't true...

Because $ab^{-1}=h$ so $hh=ab^{-1}h$

 

[micromass]

Because $ab^{-1}=h$ so $hh=ab^{-1}h$

but h isn't $ab^{-1}$.... Where did you get that??

 

[ArcanaNoir]

the very beginning, $ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h$

 

[micromass]

the very beginning, $ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h$

But that was a different proof. That was notation in order to prove (1)=> (2). You can't use that now.

Right now you're proving (3)=> (4). So to do that, you chose $ah\in aH$. All you know about h now is that it is in H!!
You're doing something different than in the beginning, so you can't use the notation from previous proofs...

 

[ArcanaNoir]

okay I'll pick a different h but I disagree about calling it a "previous proof" since the whole point of this exercise is to make the thing a connected circle.
so thats why I need something to follow directly from a \in Hb.

but first it's good to break it down into just 3 => 4
so, $h \in H \Rightarrow ha \in Ha$
Now, choosing $h' : h'=hab^{-1}$ is in H because we assumed $ab^-1 \in H$
$ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow h' \in Hh$ crap I seem lost again.

 

[ArcanaNoir]

I'm breaking out the chocolate fudge brownie ice cream now. Are you trying to make me cry? I have a lot of other proofs to study before test Tuesday :(

 

[micromass]

okay I'll pick a different h but I disagree about calling it a "previous proof" since the whole point of this exercise is to make the thing a connected circle.

You need to do 4 proofs here: (1)=> (2), (2)=> (3), (3)=> (4) and (4)=> (1). These are 4 distinct things to prove. You can not prove them simultaniously by one long chain of statements. You need to prove all four things separately.

so thats why I need something to follow directly from a \in Hb.

but first it's good to break it down into just 3 => 4
so, $h \in H \Rightarrow ha \in Ha$
Now, choosing $h' : h'=hab^{-1}$ is in H because we assumed $ab^-1 \in H$
$ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow h' \in Hh$ crap I seem lost again.

Indeed, so h' is in H. So ha=hb' is in Hb, what we needed to prove!!

 

[micromass]

I'm breaking out the chocolate fudge brownie ice cream now. Are you trying to make me cry? I have a lot of other proofs to study before test Tuesday :(

I'm sorry, I was only trying to help

I'll let somebody else handle the thread then...

 

[ArcanaNoir]

Indeed, so h' is in H. So ha=hb' is in Hb, what we needed to prove!!

what? Where?
$ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow hab^{-1} \in H$ but how does that show $ha=hb^{-1}$ ?

 

[ArcanaNoir]

I'm sorry, I was only trying to help

I'll let somebody else handle the thread then...

Sorry, I didn't mean to hurt your feelings, I only meant I was feeling dumb and frustrated.

 

[micromass]

You didn't hurt my feelings. But you have the right on the best help you can get

Anyway, you start from $ha\in Ha$. You notice that $ha=h(ab^{-1})b$ and you see that $ab^{-1}\in H$. Doesn't that prove that $ha\in Ha$??
So that shows that $Ha\subseteq Hb$. Now you need to show the converse.

 

[ArcanaNoir]

how do we know that ha is in H? we know h is in H, but we don't know that a is in H, do we?

 

[micromass]

how do we know that ha is in H? we know h is in H, but we don't know that a is in H, do we?

Sorry, typo. I fixed it.

 

[ArcanaNoir]

Does $Ha \subseteq Hb \Rightarrow a=b$? I don't think it can but somehow seems logical...
no, I got something better. (working...)

 

[micromass]

Does $Ha \subseteq Hb \Rightarrow a=b$? I don't think it can but somehow seems logical...
no, I got something better. (working...)

No, not at all. Even Ha=Hb doesn't imply a=b.

 

[ArcanaNoir]

$Ha \subseteq Hb \Rightarrow \forall h' \in H, \exists h'' in H:h'a=h''b \Rightarrow$

 

[ArcanaNoir]

Okay, giving up for now. Time for stress-free bed-time activities. :) Thanks for working on this with me micro. Sorry I'm being dense about it.

 

[Deveno]

did we ever get to 3)=>4)?

given: a is in Hb, and H is a subgroup.

showing Ha is contained in Hb is the easy part.

we pick a typical element in Ha, call it ha (so h is just some generic element of H).

now, since a is in Hb, a = h'b for some specific (but unspecified) element h' of H.

so ha = h(h'b) = (hh')b. since H is a subgroup hh' (for any h and the specific h'), is in H,

so (hh')b is "some" element of Hb.

does this show Ha is contained in Hb? (get back to me on this, what do you think?).

the tricky part is showing Hb is contained in Ha. again, we should start with some typical element of Hb. i'd say hb, but....let's just use h"b, just for clarity, to emphasize we don't mean the same "h" as in ha that we used earlier.

so h"b =....hmm. we need to use a = h'b to get something with just b on the right-hand-side. maybe multiply by an inverse? can you think which inverse we might use?

then use h"b = h"(expression for b) = (something something)a.

if the "something something" happens to be in H, that will do the trick, eh?

 

[I like Serena]

Hi Arcana!

$Ha \subseteq Hb \Rightarrow \forall h' \in H, \exists h'' in H:h'a=h''b \Rightarrow$

Slight modification for (iv)->(i):

$Ha = Hb \Rightarrow \forall h' \in H, \exists h'' \in H:h'a=h''b \Rightarrow \quad ... \quad \Rightarrow ab^{-1} \in H$

That is, start with the condition of (iv), reformulate it as specific elements, and rewrite it to the condition of (i).