[Math] Are all normal subgroups Abelian

group-theorynormal-subgroups

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?

Best Answer

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

  1. for every $h\in H$, there exists $h_1\in H$ with $xh=h_1x$
  2. 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).

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