Looking for a nice example for normal subgroups

Question

I am looking at normal subgroups at the moment and I came across a note that says that for a normal subgroup H of a group G, it is not always the case that for $a \in G$ and $h \in H$, we have $ha=ah$ for all a and h. However, it does imply that for each $h', \in H$, there is an $h'' \in H$ (where h' and h'' may be distinct) such that $h'a=ah''$.

I think that I understand what this is saying but could I see a nice example of this? I greatly appreciate any help!

Answer 1

Let us investigate some general examples.

• Let $G$ be an abelian group, then every subgroup of $G$ is normal.

• Let $G$ be a group, then $\{1_G\}$ and $G$ are normal subgroups of $G$.

• Let $G$ be a group, then the center of $G$ i.e. $\{x\in G\textrm{ s.t. }\forall g\in G,gx=xg\}$ is a normal subgroup of $G$. This is a very important example!

• Let $G$ be a group, then the commutator subgroup of $G$ i.e. the subgroup generated by $\{xyx^{-1}y^{-1},x\in G,y\in G\}$ is a normal subgroup of $G$.

Here a criterion to check whether or not a subgroup is normal.

Proposition. Let $G$ be a group and let $H$ be a subgroup of $G$, then $H$ is normal in $G$ if and only if there exists a group homomorphism $\pi:G\rightarrow G'$ such that $H=\ker(\pi)$.

Proof. I let the backward implication to you, it is only a verification and it is a good exercise.

Assume $H$ is normal, then $G/H$ is a group and the projection $\pi:G\twoheadrightarrow G/H$ is a group homomorphism whose kernel is $H$. $\Box$

I would like to conclude with the following proposition:

Proposition. Let $G$ be a finite group and let $p$ be the smallest divisor of $|G|$, then every subgroup of index $p$ is normal.

Proof. Let $H$ be a subgroup of $G$ of index $p$, then $H$ acts on $G/H$ by left translations. Let $G/H^H$ be the set of elements of $G/H$ fixed by any elements of $H$. Notice that $H$ is a normal if and only if $G/H^H=G/H$. It is easy to see that $H\in G/H^H$, hence $|G/H^H|\geqslant 1$. Assume there exists $gH\in(G/H)\setminus(G/H^G)$, using orbit-stabilizer lemma, one has: $$|\omega(gH)|\times|\textrm{Stab}(gH)|=|H|.$$ Therefore, $|\omega(gH)|\geqslant p$, which is a contradiction since $G/H$ is a disjoint union of $\omega(gH),g\in G$. $\Box$

Answer 2

Typical explicit examples are:

1.) The subgroup $A_n$ of $S_n$ is normal (since it has index $2$).

2.) The subgroup of translation $T(n)$ of $\mathbb{R}^n$ is a normal subgroup of the isometry group ${\rm Isom}(\mathbb{R}^n)$.

3.) The subgroup of rotations $C_n$ in the dihedral group $D_n$ is normal (since it has index $2$).

4.) The group $SL_n(\mathbb{Z})$ is a normal subgroup of $GL_n(\mathbb{Z})$ (as a kernel of a group homomorphism).

5.) Every subgroup of the non-abelian group $Q_8$, the quaternion group, is normal - see here.