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Are these lines logical?

  • #1
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To make a very long story short, in a group theory problem I am working on, I need to prove this:

##A \lhd B \Rightarrow A'\neq A##,
where ##A## and ##B## are finite and ##A'## is called the commutator subgroup:

##\begin{align}
A' :&= [A, A] \\
&= \langle [x, y] \mid x, y \in A \rangle \\
&= \langle x^{-1}y^{-1}xy \rangle
\end{align}##.
Here are the lines I made out: Since ##A## is a normal subgroup, therefore it is commutative. For ##\forall x, \forall y \in A##,

##\begin{align}
xy &= yx && (1)\\
y^{-1}xy &= x && (2)\\
y^{-1}xy &\in A && (3)\\
x^{-1}y^{-1}xy &\notin A && (4)\\
\langle x^{-1}y^{-1}xy \rangle &\neq A && (5)\\
\langle [x, y] \rangle &\neq A && (6)\\
[A, A] &\neq A && (7)\\
A' &\neq A && (8)
\end{align}##.​

With this posting, I am very interested to know if the logic from line (3) to (4) and to (5) are valid, and if they are not, I would love to learn how they should be instead. Thank you for your time and help.
 

Answers and Replies

  • #2
haruspex
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I can think of no basis at all for your step (3) to (4). Indeed, isn't the term on the left in (4) the identity element?
Since A is a normal subgroup, therefore it is commutative.
Are you sure? A long time since I studied groups, but that doesn't sound familiar.
 
  • #3
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I can think of no basis at all for your step (3) to (4). Indeed, isn't the term on the left in (4) the identity element?
Honestly I am so doubtful of this logic that I decided to put it up here as a question. Do you have any other suggestion then?
 
  • #4
haruspex
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Honestly I am so doubtful of this logic that I decided to put it up here as a question. Do you have any other suggestion then?
Why do you think what you are trying to prove is even true?
Isn't the trivial group consisting of the identity {1} a normal subgroup of B? And is it not equal to its own commutator subgroup?
 
  • #5
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Are you sure? A long time since I studied groups, but that doesn't sound familiar.
What I am trying to imply is that since the definition of normal subgroup goes like this:
##N \lhd G \Leftrightarrow \forall n \in N, \forall g \in G, ng = gn##,​
does this imply that ##n## is then abelian? Let me know and thanks.
 
  • #6
haruspex
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What I am trying to imply is that since the definition of normal subgroup goes like this:
##N \lhd G \Leftrightarrow \forall n \in N, \forall g \in G, ng = gn##,​
does this imply that ##n## is then abelian? Let me know and thanks.
No, that's not the definition of a normal subgroup. g-1ng is an element of N, but not necessarily n.
http://en.wikipedia.org/wiki/Normal_subgroup
 
  • #7
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Why do you think what you are trying to prove is even true?
Isn't the trivial group consisting of the identity {1} a normal subgroup of B? And is it not equal to its own commutator subgroup?
Let me type in the whole question so that you can see it as a big picture, give me about 5 minutes, if you will. Thanks.
 
  • #8
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Why do you think what you are trying to prove is even true?
Isn't the trivial group consisting of the identity {1} a normal subgroup of B? And is it not equal to its own commutator subgroup?
I typed in the original question couple of days on MSE here, but I did not get any help perhaps because I typed in the question way TOO LONG, therefore nobody gets interested and therefore it is lost forever, and there is no way to resuscitate it back from dead. I would love to get any idea from you. Thanks.
 
  • #9
haruspex
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Ok. In this thread you omitted the key constraint that A is not simply {1}.
Can you show that if Hi/Hi-1 is abelian then Hi-1 = Hi'?
 
  • #10
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Ok. In this thread you omitted the key constraint that A is not simply {1}.
Can you show that if Hi/Hi-1 is abelian then Hi-1 = Hi'?
Is this the direction you are hinting me to pursue: If I can prove that ##H_{i-1} = H'_i##, then it implies ##H_i \neq H'_i##?
 
  • #11
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Is this the direction you are hinting me to pursue: If I can prove that ##H_{i-1} = H'_i##, then it implies ##H_i \neq H'_i##?
No, that would be trivial. ##H_{i-1} \neq H_i## by definition, so if ##H_{i-1} = H'_i##, then ##H_i \neq H'_i##.
 
  • #12
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Ok. In this thread you omitted the key constraint that A is not simply {1}.
Can you show that if Hi/Hi-1 is abelian then Hi-1 = Hi'?
Ok, here is what I understand of factor group: If ##b \in H_i## then the factor group ##H_i / H_{i-1} = \{ bH_{i-1} \}##, am I correct? And then since ##H_i / H_{i-1}## is abelian, therefore ##bH_{i-1} = H_{i-1}b##? Am I correct?
 
  • #13
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  • #14
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  • #15
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I don't see why. It certainly implies ##bH_{i-1} aH_{i-1} = aH_{i-1} bH_{i-1}## for all a, b in ##H_i##.
Take a look at https://www.proofwiki.org/wiki/Abelian_Quotient_Group
Are you still there? I am continuing the conversation, suppose that ##N## is normal subgroup of ##G## and ##G/H## is abelian:

##\begin{align}
\forall aH,bH &\in G/H \\
aHbH &= bHaH \qquad \qquad &&(1) \\
abH &= baH &&(2) \\
ab(ba)^{−1} &\in H &&(3a) \\
ab(ba)^{−1} &=H &&(3b) \\
\ldots & \ldots \\
aba^{−1}b^{−1} &= H &&(4)\\
[a, b] &= H &&(5)\\
G' &= H &&(6) \\
\end{align} ##
My question is: Which one is correct, the (3a) or (3b)? As you advise, my final goal is ##G' = H##. Please advise and thanks again.

 
  • #16
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Are you still there? I am continuing the conversation, suppose that ##N## is normal subgroup of ##G## and ##G/H## is abelian:

##\begin{align}
\forall aH,bH &\in G/H \\
aHbH &= bHaH \qquad \qquad &&(1) \\
abH &= baH &&(2) \\
ab(ba)^{−1} &\in H &&(3a) \\
ab(ba)^{−1} &=H &&(3b) \\
\ldots & \ldots \\
aba^{−1}b^{−1} &= H &&(4)\\
[a, b] &= H &&(5)\\
G' &= H &&(6) \\
\end{align} ##
My question is: Which one is correct, the (3a) or (3b)? As you advise, my final goal is ##G' = H##. Please advise and thanks again.

I assume you mean H, not N.
Clearly 3(b) is not right, but maybe you meant ##<ab(ba)^{−1}> =H##.
My reading of the link I posted is that you can only say ##<ab(ba)^{−1}> \leq H##.
But maybe that's enough to get to the result you are after.
 
  • #17
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Thank you for your time and help.
 

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