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 that's 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 that's 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).

 

[Deveno]

Hi Arcana!



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).

what i like to do is think of these things like this:

Ha = Hb. so now we need something "element-wise" involving a and b.

so every h'a in Ha is some h"b in Hb. don't really care which, we're not peeking that far under the hood, so we just pick some h' and h" that makes that true:

h'a = h"b.

now we need to get to something that is of the form:

blah blah blah = ab^-1

to get a all by itself, we want to multiply the equation by something on the left, or otherwise we'll mix up h-somethings with the b (that would be counter-productive). h'^-1 seems like the obvious choice:

a = h'^-1h"b.

now...do you see how to get b^-1 on the left, and take b off the right?

 

[ArcanaNoir]

I'm not going to continue working on this problem.

 

[I like Serena]

Must be something in the air. Perhaps autumn?
You're the second today that's apparently frustrated and says so in the thread.

Ah wait! Today daylight saving switched to wintertime.
Did your clock switch?

 

[Deveno]

well, that's your choice, but:

if at some later date, some theorem uses this fact without further comment in its proof (and i can assure you that will happen), you will find yourself mystified.

these 4 conditions are used interchangeably to identify common cosets of H, and crop up again and again. this problem isn't just "some example to show how groups work", it's part of the "standard vocabulary" of cosets.

so, skipping this problem, it may not negatively affect, say, your grade on this assignment. but it's like deciding "i'm not going to learn how to read words that start with the letter T".

math education is cumulative, if you skip some small parts early on, you have "bigger holes" later.

i mean, look, it's no skin off MY nose if you work through this or not, but i am telling you straight-up, you're selling yourself short. expect a steep uphill climb in the rest of the course.

 

[ArcanaNoir]

well, that's your choice, but:

if at some later date, some theorem uses this fact without further comment in its proof (and i can assure you that will happen), you will find yourself mystified.

these 4 conditions are used interchangeably to identify common cosets of H, and crop up again and again. this problem isn't just "some example to show how groups work", it's part of the "standard vocabulary" of cosets.

so, skipping this problem, it may not negatively affect, say, your grade on this assignment. but it's like deciding "i'm not going to learn how to read words that start with the letter T".

math education is cumulative, if you skip some small parts early on, you have "bigger holes" later.

i mean, look, it's no skin off MY nose if you work through this or not, but i am telling you straight-up, you're selling yourself short. expect a steep uphill climb in the rest of the course.

Gee, is everyone cranky today?
Just because I'm not doing this problem at this time doesn't mean I won't ever learn it. It's just that I have a huge test on Tuesday and many many theorems and exercises to be familiar with and this one is just taking too much time and frustrating me. I'm past the point where I can mentally deal with this problem because my stress level is just too high right now. I do appreciate all the help I've received on this problem, but if I'm going to postpone having a mental breakdown until after the test, then this problem needs to be put on the shelf for now. Yes, it's entirely possible this problem might be on this test, but it also might not, and so in essence I'm sacrificing my potential to get this problem on the test for the greater good of studying several other theorems that are equally likely to be on the test. :) (yes that's the smile of a crazy person trying grasp at the straws of sanity)

 

[Deveno]

Gee, is everyone cranky today?
Just because I'm not doing this problem at this time doesn't mean I won't ever learn it. It's just that I have a huge test on Tuesday and many many theorems and exercises to be familiar with and this one is just taking too much time and frustrating me. I'm past the point where I can mentally deal with this problem because my stress level is just too high right now. I do appreciate all the help I've received on this problem, but if I'm going to postpone having a mental breakdown until after the test, then this problem needs to be put on the shelf for now. Yes, it's entirely possible this problem might be on this test, but it also might not, and so in essence I'm sacrificing my potential to get this problem on the test for the greater good of studying several other theorems that are equally likely to be on the test. :) (yes that's the smile of a crazy person trying grasp at the straws of sanity)

i understand you are stressed out because you have a lot of material to cover, and you don't want to spend a lot of time on "just one problem" that's driving you crazy. and i don't know how much material (how many chapters, concepts, etc.) is going to be covered on this test.

what i would like to point out, is that not all problems are created equal. some are more "disposable" than others. and what I'm trying to communicate to you is that THIS problem, is actually important, you will use this later on (even if it's NOT an important question on your test). i can't tell you if "getting" this exercise wil help you on the test, or not, it may not. but what i CAN say, with certainty, is that if you do not understand this problem inside-out, you will struggle with cosets for...well, at least as long as you're doing group theory.

many textbooks (Dummit and Foote, Herstein, Jacobsen, Artin) will prove something like ab^-1 is in H, and then use the cosets Ha and Hb interchangeably. at some point, you will probably cover the orbit-stabilizer theorem, and this entire problem will be "condensed" in it.

i'm not trying to be an unsymapthetic person. i know what it is like to have x amount of time, and x + y amount of stuff to review (where y is large compared to x). what i am trying to give you some sense of, is that regardless of how well you do on "the test", you need to know this stuff. the object of taking a course, is to acquire the knowledge, not "a good grade in it".

stumbling over basic concepts, in order to do well short-term, isn't a good long-term strategy. you may do fine on the upcoming test, by ignoring this issue, and covering other material you've been ignoring, and i understand the value of that to you. but, in the long-run, i'd rather see you ace the final, or the mid-term, even at the expense of some stuff in-between, rather than start out strong, and get hopelessly stuck at the end.

so...look, i can't tell you what to do, and in the end, it's not really my business. but...if you don't do this sometime, you'll never really LEARN this stuff, and then whatever grade you got, is meaningless, it's just some printing on some paper somewhere.

so...here's a compromise...let it sit a bit, you're obviously tired of banging your head against this wall. take a quick look sometime tomorrow, maybe it will make some more sense after you've relieved some of that stress.

 

[Evo]

The Op has requested a rest for the thread.