Exercise 3.1.3 (Examples of non abelian groups $G$ such that $G/N$ is abelian)

Question

Let $A$ be an abelian group and let $B$ be a subgroup of $A$. prove that $A/B$ is abelian. Give an example of a non-abelian group $G$ containing a proper normal subgroup $N$ such that $G/N$ is abelian.

Richard Ganaye · Accepted Answer

\begin{proof} 
Since $A$ is abelian, every subgroup of $A$ is normal in $A$. Let $\overline{a} = a B$ and $\overline{b} = b B$ be any elements of $A/B$. By definition of the law of $A/B$,
$$\overline{a}\, \overline{b} = aB\, bB = ab B = ba B = bB\, a B = \overline{b}\, \overline{a}.$$
Therefore $A/B$ is an abelian group.

\bigskip
Consider the non abelian group $G = D_8$ and its center $N = Z =\{1, r^2\}$, which is normal is $D_8$. Then $D_8/Z = \{\overline{1}, \overline{r}, \overline{s}, \overline{rs}\}$ has 4 elements (see Example (3)). Therefore $G/N = D_8/Z$ is abelian (Exercise 2.5.10).

Another less easy example is the quotient $M/\langle v^4 angle \simeq Z_2 	imes Z_4$, where $M$ is the modular group of order $16$ (see the solution of Exercise 2.5.15).
\end{proof}

Exercise 3.1.3 (Examples of non abelian groups $G$ such that $G/N$ is abelian)

Answers

Comments

Exercise 3.1.3 (Examples of non abelian groups $G$ such that $G/N$ is abelian)

Answers

Comments

Add answer