Proof involving group and subgroup

  • Thread starter ArcanaNoir
  • Start date
  • #1
768
4

Homework Statement


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


Homework Equations


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


The Attempt at a Solution



[itex] 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 ...? [/itex]
 

Answers and Replies

  • #2
22,089
3,294
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.
 
  • #4
22,089
3,294
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'.
 
  • #5
768
4
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 [itex] h^{-1}=ba^{-1} [/itex] so [itex] ba^{-1}=h^{-1} [/itex] and [itex] h=ab^{-1} \Rightarrow [/itex]
But I'm not getting anywhere. What should come directly after [itex] a \in Hb \Rightarrow [/itex] ?
 
  • #6
22,089
3,294
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 [itex]h^\prime=b^{-1}ah[/itex]. But why is this in H?
 
  • #7
768
4
is it a problem that I'm looking for Ha, not aH?
 
  • #8
22,089
3,294
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 [itex]h^\prime=hab^{-1}[/itex]. but why is this in H?
 
  • #9
768
4
okay, h is in H already, so hh is in H, and [itex] hh=hab^{-1}=h' \in H [/itex]
 
  • #10
768
4
so [itex] h' \in H \Rightarrow h'b \in Hb \Rightarrow ah'b \in Hb \Rightarrow ah' \in H \Rightarrow [/itex]
[itex] h^{-1}ah' \in H \Rightarrow ba^{-1}ah' \in H \Rightarrow bh' \in H [/itex]
 
Last edited:
  • #12
768
4
Why?? This isn't true...
Because [itex] ab^{-1}=h [/itex] so [itex] hh=ab^{-1}h [/itex]
 
  • #13
22,089
3,294
Because [itex] ab^{-1}=h [/itex] so [itex] hh=ab^{-1}h [/itex]
but h isn't [itex]ab^{-1}[/itex].... Where did you get that??
 
  • #14
768
4
the very beginning, [itex] ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h [/itex]
 
  • #15
22,089
3,294
the very beginning, [itex] ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h [/itex]
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 [itex]ah\in aH[/itex]. 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...
 
  • #16
768
4
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, [itex] h \in H \Rightarrow ha \in Ha [/itex]
Now, choosing [itex] h' : h'=hab^{-1} [/itex] is in H because we assumed [itex] ab^-1 \in H [/itex]
[itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow h' \in Hh [/itex] crap I seem lost again.
 
  • #17
768
4
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 :(
 
  • #18
22,089
3,294
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, [itex] h \in H \Rightarrow ha \in Ha [/itex]
Now, choosing [itex] h' : h'=hab^{-1} [/itex] is in H because we assumed [itex] ab^-1 \in H [/itex]
[itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow h' \in Hh [/itex] crap I seem lost again.
Indeed, so h' is in H. So ha=hb' is in Hb, what we needed to prove!!
 
  • #19
22,089
3,294
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 :frown:

I'll let somebody else handle the thread then... :frown:
 
  • #20
768
4
Indeed, so h' is in H. So ha=hb' is in Hb, what we needed to prove!!
what? Where?
[itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow hab^{-1} \in H [/itex] but how does that show [itex] ha=hb^{-1} [/itex] ?
 
  • #21
768
4
I'm sorry, I was only trying to help :frown:

I'll let somebody else handle the thread then... :frown:
Sorry, I didn't mean to hurt your feelings, I only meant I was feeling dumb and frustrated.
 
  • #22
22,089
3,294
You didn't hurt my feelings. But you have the right on the best help you can get :smile:

Anyway, you start from [itex]ha\in Ha[/itex]. You notice that [itex]ha=h(ab^{-1})b[/itex] and you see that [itex]ab^{-1}\in H[/itex]. Doesn't that prove that [itex]ha\in Ha[/itex]??
So that shows that [itex]Ha\subseteq Hb[/itex]. Now you need to show the converse.
 
Last edited:
  • #23
768
4
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?
 
  • #24
22,089
3,294
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.
 
  • #25
768
4
Does [itex] Ha \subseteq Hb \Rightarrow a=b [/itex]? I don't think it can but somehow seems logical...
no, I got something better. (working...)
 

Related Threads on Proof involving group and subgroup

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
17
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
19
Views
2K
Top