- #1

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##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.