Are all normal subgroups Abelian? If $H \subset G$ is a normal subgroup of G,
=> $xHx^{-1} = H$ or $xH = Hx$ for all $x \epsilon G$
=> $xH = Hx$ for all $x \epsilon H$
Hence, all normal subgroups of a group are themselves Abelian?
Also, does that mean that all normal towers of subgroups are also Abelian?
 A: No. $G$ is a normal subgroup of itself for any group $G$. Commutativity is a local property of a group (all the way down to the elements), whereas normality is more of a mid-scale property of a group (collections of elements, but not necessarily the whole group). Just because you permute the elements around the same way if you multiply on the left and right (in regards to normality) does not mean it is abelian.
A: It is quite a common misunderstanding that $xH=Hx$ means that $xh=hx$ for any $h\in H$. This is generally false.
The assertion $xH=Hx$ means that


*

*for every $h\in H$, there exists $h_1\in H$ with $xh=h_1x$

*for every $h\in H$, there exists $h_2\in H$ with $hx=xh_2$


Consider the group $G=S_4$ and its normal subgroup $H=A_4$ (the even permutations). Since $[S_4:A_4]=2$, the subgroup $A_4$ is normal. However, taking $x=(12)$ and $h=(123)$, we have
$$
(12)(123)=(23)\ne(123)(12)=(13)
$$
However, $(132)(12)=(23)$, so in this case $h_1=(132)\ne h$.
We can also find two elements in $A_4$ that don't commute:
$$
(123)(124)=(13)(24)
\\
(124)(123)=(14)(23)
$$
Note: the convention about function composition is the standard functional one, that is, on the left (think to $\circ$ between two cycles).
A: To get other counterexamples, $A$ and $B$ are normal subgroups in the direct product $A \times B$.
