# Are these lines logical?

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

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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.

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?

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?

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.

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

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.

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.

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'?

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 $H_i \neq H'_i$?

haruspex
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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 $H_i \neq H'_i$.

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?

haruspex
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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
Thanks! I think I am about to get it now, I will keep working on it. Thanks again and again.

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.

haruspex
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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.

Thank you for your time and help.