- #1
A.Magnus
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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:
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.
##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##,A' :&= [A, A] \\
&= \langle [x, y] \mid x, y \in A \rangle \\
&= \langle x^{-1}y^{-1}xy \rangle
\end{align}##.
##\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}##.
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.