(a) Let $H$ be a normal subgroup of $G$. If $H$ are $G/H$ are Abelian, should $G$ be abelian?
Attempt: : There's a counterexample to this claim, $G=D_3$ which is non abelian.
But, what could be wrong in this procedure :
$G/H = \{gH~\forall~g \in G\}$
$G/H$ is abelian means $aH ~bH = bH ~aH$
now, $aH ~bH = H abH$ since $H$ is a normal subgroup, which will be $= bH ~aH$ only when $a$ and $b$ commute. Since, $a,b$ represent the general elements of $G$, shouldn't $G$ be abelian?
(b) Let $G= Z_4 \oplus U(4)$, $H =\langle(2,3) \rangle , K=\langle (2,1) \rangle$ . Prove that $G/H \not\approx G/K$
Attempt: $G/H = \{g H ~\forall ~g \in G\} = (z_1+2 \mod 4, ~3~u_1 \mod 4)$
$G/K = \{g K ~\forall ~g \in G\} = (z_2+2 \mod 4,~ u_2 \mod 4)$
where $z_1,z_2 \in Z_4, u_1,u_2 \in U(4)$
How do I proceed next?
Help will be appreciated. Thank you
(a) $HabH=HbaH$ does not imply $ab=ba$. The counterexample $D_3$ is fine.
(b) The procedure might depend on the undefined $K$.