Are all normal subgroups Abelian?

Question

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?

An easy counterexample is $A_4 \trianglelefteq S_4$ but $A_4$ is not abelian. — lokodiz, May 03 '16 at 21:27
One thing it think about is that H is a set, so it may be confusing if you are thinking of elements in G or H. — wesssg, May 04 '16 at 03:26

egreg · Accepted Answer · 2017-06-22T08:53:36.157

27

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).

edited Jun 22 '17 at 08:53

answered May 03 '16 at 21:38

egreg

238,574

That is exactly what I thought, but then realized that $xyx^{-1} = z$ for $y,z \epsilon H$ does not imply $y = z$, rather $xyx^{-1} \epsilon H$ – ixaxaar May 03 '16 at 21:42
What notation is that? It doesn't seem to work if it's permutation cycle notation. – Mateen Ulhaq Jun 22 '17 at 08:39
@MateenUlhaq I use composition in the standard direction, so $f=(123)$, $g=(124)$ and $f\circ g(1)=f(2)=3$ and so on. – egreg Jun 22 '17 at 08:42
@egreg Ah OK, thanks. Is that a typical convention in group theory? – Mateen Ulhaq Jun 22 '17 at 08:48
1

@MateenUlhaq Depending on the textbook it can be reversed. – egreg Jun 22 '17 at 08:49
@egreg However, any subgroup of an Abelian group is normal since if $G$ is abelian then for any $x\in X$ the equality $$xh=hx$$ with $h\in H$ holds, right? – Antonio Maria Di Mauro Mar 07 '23 at 17:38
@AntonioMariaDiMauro Yes, of course. – egreg Mar 07 '23 at 21:23
@egreg Perfect: thanks for the answer! – Antonio Maria Di Mauro Mar 07 '23 at 21:45
1

@AntonioMariaDiMauro On the other hand, there exist nonabelian groups such that every subgroup thereof is normal. Be careful! – egreg Mar 07 '23 at 22:50
@egreg Oh, nice! I did not know it. – Antonio Maria Di Mauro Mar 08 '23 at 08:40

score 11 · Answer 2 · answered May 03 '16 at 21:20

11

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.

answered May 03 '16 at 21:20

Cameron Williams

29,432

Got it, thanks! – ixaxaar May 03 '16 at 21:33

score 2 · Answer 3 · answered May 04 '16 at 07:14

2

To get other counterexamples, $A$ and $B$ are normal subgroups in the direct product $A \times B$.

answered May 04 '16 at 07:14

Seirios

33,157

Are all normal subgroups Abelian?

3 Answers3

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