Dihedral Group D4/Normal Subgroups/Subgroup Generated by a^2/Quotient Group/Subgroups

From ProofWiki
Jump to navigation Jump to search

Quotient Group of Normal Subgroup of the Dihedral Group $D_4$

Let the dihedral group $D_4$ be represented by its group presentation:

$D_4 = \gen {a, b: a^4 = b^2 = e, a b = b a^{-1} }$

Consider the quotient group of the normal subgroup $\gen {a^2}$ of $D_4$:

$G / N = \set {E, A, B, C}$


$E := \set {e, a^2}, A := a \set {a, a^3}, B := b \set {b, b a^2}, C := a b \set {b a, b a^3}$.

The subgroups of $G / N$ are:

\(\ds E\) \(=\) \(\ds \set {e, a^2}\)
\(\ds \set {E, A}\) \(=\) \(\ds \set {e, a, a^2, a^3}\)
\(\ds \set {E, B}\) \(=\) \(\ds \set {e, b, a^2, b a^2}\)
\(\ds \set {E, C}\) \(=\) \(\ds \set {e, b a, a^2, b a^3}\)
\(\ds \set {E, A, B, C}\) \(=\) \(\ds D_4\)

As $G / N$ is abelian, all of these subgroups are normal in $G / N$.

Also, all of these are normal in $D_4$:

$E$ is normal a priori
$\set {E, A}, \set {E, B}, \set {E, C}$ are all of index $2$, and so normal by Subgroup of Index 2 is Normal.
$D_4$ is normal by Group is Normal in Itself.