Index of subgroup H is 2 implies

[ArcanaNoir]

Index of subgroup H is 2 implies...

Homework Statement

Just had my abstract algebra test. This was the only question I did not answer. The rest I answered somewhat confidently.

Prove that if H is a subgroup of G, [G : H] = 2, a,b are in G but not in H, then ab is in H.

Homework Equations

2 is the index of H, that is, the order of G equals 2 times the order of H.

The Attempt at a Solution

I have no idea where to start with this. I can't even come up with a concrete example demonstrating this.

 

[kru_]

The concrete example that springs to mind is G = Z_6 and H is the subgroup of G generated by 2. Then 1 and 3 are in G and 1+3=4 is in H.

I am not sure how to help prove the general case, but I am interested in the proof.

 

[ArcanaNoir]

Hmm, good example! I wonder if this only works with cyclic groups.

 

[I like Serena]

Hi Arcana!

I take it congratulations are in order, if you were so confident?Another non-cyclic example would be the 4-group of Klein: de symmetries of the diamond.
This is G={id,sx,sy,h}
If H={id,sx} then with every choice for a and b from {sy,h} the product will be in H.
Now for a proof.
I'll show you how it starts...

The index [G] is the order of the quotient group G/H.
If it is 2, then there are 2 elements in G/H.
Let's say G/H={H, cH}.

Now every a and b that are not in H must therefore be in cH, which is also aH then.
So there is an h' in H for which b=ah'

Suppose ab is also not in H.
Then there is an h'' in H for which ab=ah''.

...

 

[Deveno]

no, it works with any group with a subgroup of index 2.

if we have a subgroup of index 2, we have just two cosets of H, H and "the rest of G".

one consequence of this, is that for any g in G, gH = Hg.

for a,b not in H, consider the set S = {hah'b : h,h' in H}.

note that since ah' is in aH, ah' = h"a, for some other element h" of H.

thus S is contained in Hab (hah'b = hh"ab). on the other hand,

for any element hab in Hab, we can write hab = (ha)(eb), which is in S.

so S and Hab are the same set.

since S is the coset Hab, we have 2 choices: S = H, or S = Ha = Hb (those are the only 2 cosets we have).

suppose b is in S = Hab (this is the choice S = Hb).

then b = hab, so e = ha, which means e is in Ha,

which means that H = Ha, so that a is in H. but a isn't in H, so this is a contradiction.

why does this show ab is in H?

*****

note to I like Serena: we're saying the same thing. what i do not know, is whether or not they've covered normality yet, so i have fastidiously avoided the term "quotient group", i am only dealing with a set of (right) cosets, and i only use the property gH = Hg once, to avoid such a digression.

 

[ArcanaNoir]

Why is there an h in H such that b=ah?

(class is just now ending, I am heading for home, so I won't see anymore replies tonight)

 

[I like Serena]

Why is there an h in H such that b=ah?

(class is just now ending, I am heading for home, so I won't see anymore replies tonight)

Let's illustrate, suppose H={e,h₁,h₂,h₃}
And G/H={H, cH}.
This is the case because the index [G] is defined to be the number of elements in G/H, which is 2 in this case.
Then cH={c,ch₁,ch₂,ch₃}
And therefore G={e,h₁,h₂,h₃, c,ch₁,ch₂,ch₃}.There are 2 elements in G/H, H itself and some cH.
So each element in G must be either in H or in cH.

b is not in H, so b must be in cH=aH.
Therefore there must be a h in H, such that b=ah.

 

[ArcanaNoir]

Okay, so b must equal c, ch₁, ch₂, or ch₃.
Also, so must a.

so maybe b=ch₁ and a=ch₂
Then if ab is in H, that must mean ch₂ch₁ is in H. So how do we reduce ch₂ch₁ to e, h₁, h₂, or h₃?

 

[I like Serena]

Okay, so b must equal c, ch₁, ch₂, or ch₃.
Also, so must a.

so maybe b=ch₁ and a=ch₂
Then if ab is in H, that must mean ch₂ch₁ is in H. So how do we reduce ch₂ch₁ to e, h₁, h₂, or h₃?

Well typically you would do this with a proof by contradiction.

Suppose ch₂ch₁ is not in H.
Then ch₂ch₁ must be in cH (due to the [G]=2 index thingy).
Therefore there must be an h in H, such that ch₂ch₁ = ch.
Can you do some cancellations and stuff?

 

[ArcanaNoir]

No, this is getting a little messy. Beep beep, back up the truck. Let's try again.

a and b are not in H, thus they are in the other half of G, aka cH, aka the coset of H.

Suppose ab is not in H. Then, ab is in cH. so...?

 

[I like Serena]

No, this is getting a little messy. Beep beep, back up the truck. Let's try again.

a and b are not in H, thus they are in the other half of G, aka cH, aka the coset of H.

Suppose ab is not in H. Then, ab is in cH. so...?

The definition of cH is $\{x: x=ch \wedge h \in H\}$.
So there is an h in H, such that ab=ch.

 

[ArcanaNoir]

Okay, so then...? (sorry I'm not getting it yet)

 

[I like Serena]

Okay, so then...? (sorry I'm not getting it yet)

We picked a and b from cH, and we try to prove that ab in cH will lead to a contradiction.

Converting this to specific elements we get:

$\exists h_1, h_2, h_3 \in H \text{ such that } a=ch_1, b=ch_2, ab=ch_3$

Substitute a and b in ab, and try to find a contradiction...

 

[ArcanaNoir]

ch₁ch₂=ch₃ implies (using left cancellation law)
h₁ch₂=h₃
But then I'm stuck.

 

[I like Serena]

ch₁ch₂=ch₃ implies (using left cancellation law)
h₁ch₂=h₃
But then I'm stuck.

What happens if you left-multiply with $h_1^{-1}$ and use that H is closed under multiplication?

 

[ArcanaNoir]

then I get ch₂=h₃h₁^-1
which implies h₃h₁^-1 is in cH and not in H but it must be in H since h₃ and h₁^-1 are in H. Thus, contradiction, thus ab is in H.

right?

I do not like that problem.

 

[I like Serena]

Yep!

What's wrong with the problem?
There are a few interesting concepts in there.
Such that you can divide a group in disjunct cosets.

Visually, I like to think graphically in the X-Y plane.
A subgroup is for instance a set of points on the x-axis.
And another coset is a line parallel to the x-axis.
The operation is vector addition, but modulo some number.
Obviously if you pick 2 vectors in the parallel line, they won't sum up to a vector on that line.

Visualizing stuff makes it come a bit to life for me.

Btw, the proof can be a bit shorter and neater if you use for instance that cH=aH, effectively discarding c.
(In the vector representation of a line, you are free to choose the supporting vector.)

 

[ArcanaNoir]

thanks for the help. I am not a visual person, graphs make me sad :( I'm going to be sad a lot in vector calc next semester.

 

[I like Serena]

Ah, you are more in number theory then?
Then you should appreciate abstract algebra all the more! ;)

 

[ArcanaNoir]

yes, I love my abstract algebra class. i just didn't dig this problem. I'm not too keen on cosets.

 

[I like Serena]

Well, you're not done with cosets by a long shot! Just you wait!
(Did you already skip ahead in your algebra material? )

 

[ArcanaNoir]

well next we do isomorphisms. after that, looks like homomorphisms, with a section on quotient groups, and then rings.

 

[I like Serena]

Well, there should be a section about the grand "Isomorphism theorem".
It may be included in the section on quotient groups, which is also about cosets.
You should get some headaches from that. (I did! )

 

[ArcanaNoir]

oh dear. "Isomorphism Theorem (parts 1 and 2)" is in the problems section! If it's the headache you say it is, I'm going to be unhappy!

 

[I like Serena]

No worries, all you have to do is apply Mike's Mathgasm cycle!

 

[ArcanaNoir]

Lol!