Group Theory 9: Normal Subgroups

In here, it was shown that the kernel $K=\ker\varphi$ of a homomorphism $\varphi: G\longrightarrow G’$ is a subgroup of $G$ and that it satisfies the property $aK=Ka$ for all $a\in G$. The kernel of a homomorphism is an example of a particular kind of subgroups of a group called normal subgroups.

Definition. The subgroup $N$ of $G$ is called a normal subgroup of $G$ if $a^{-1}Na\subset N$, $\forall a\in G$. If $N$ is a normal subgroup of $G$, we write $N\vartriangleleft G$.

Note that $\forall a\in G$, $a^{-1}Na\subset N$ implies that $\forall a\in G$, $a^{-1}Na=N$. Hence, we have

Theorem. $N\vartriangleleft G$ if and only if $\forall a\in G$, $aN=Na$.

If $G$ is abelian, every subgroup is a normal subgroup. But the converse need not be true. There are nonabelian groups in which every subgroup is normal. Such nonabelian groups are called Hamiltonian.

Example. Let $\mathbb{Q}_8=\{\pm 1,\pm, i, \pm j, \pm k\}$ where $1,i,j,k$ satisfy the multiplication laws
\begin{align*}
1^2&=1,\ i^2=j^2=k^2=-1,\\
ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j.
\end{align*}
Then $\mathbb{Q}_8$ forms a nonabelian group. There are four subgroups of $\mathbb{Q}_8$:
\begin{align*}
\langle -1\rangle&=\{1,-1\},\\
\langle i\rangle&=\{1,-1,i,-i\},\\
\langle j\rangle&=\{1,-1,j,-j\},\\
\langle k\rangle&=\{1,-1,k,-k\}.
\end{align*}
All these subgroups are normal subgroups of $\mathbb{Q}_8$, so $\mathbb{Q}_8$ is Hamiltonian. Note that $1,i,j,k$ form a 4-dimensional algebra of quaternions over $\mathbb{R}$ $$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$ The real algebra of quaternions $\mathbb{H}$ is identified with the 4-dimensional Euclidean space $\mathbb{R}^4$.

Recall that the index $|G:H|$ of a subgroup $H$ in $G$ is the number of right cosets of $H$ in $G$ or equivalently the number of left cosets of $H$ in $G$ as discussed here.

Theorem. Let $G$ be a group and $N\leq G$. If $|G:N|=2$ then $N\vartriangleleft G$.

Proof. Let $a\in G$. If $a\in N$, $aN=Na$. Now suppose that $a\in G\setminus N$. Then
$$N\cap Na=\emptyset=N\cap aN.$$
Furthermore, since $G=N\cup Na=N\cup aN$, $Na=Na$.

Example. Recall that $|S_n|=n!$ and $|A_n|=\frac{n!}{2}$. Since $|S_n:A_n|=2$, $A_n\vartriangleleft S_n$.

Remark. The converse of the above theorem need not be true, namely the index $|G:N|$ of a normal subgroup $N$ in $G$ is not necessarily 2. The 4th dihedral group (we introduced it here)
$$D_4=\{1,\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau\}$$
has the following nontrivial subgroups:
\begin{align*}
&\{1,\sigma^2,\sigma\tau,\sigma^2\tau\},\ \{1,\sigma,\sigma^2,\sigma^3\},\ \{1,\sigma^2,\tau,\sigma^2\tau\},\\
&\{1,\sigma^3\tau\},\ \{1,\sigma\tau\},\ \{1,\sigma^2\},\ \{1,\tau\},\ \{1,\sigma^2\tau\}.
\end{align*}
The subgroups of order 4 are normal subgroups of $D_4$ because their indices are 2. The center of $D_4$ is $\{1,\sigma^2\}$, so it is also a normal subgroup of $D_4$. However, $|D_4:Z(D_4)|=4$. The rest of nontrivial subgroups are not normal subgroups of $D_4$.

Remark. $H\vartriangleleft N, N\vartriangleleft G\not\Longrightarrow H\vartriangleleft G$. In the previous remark, $H=\{1,\tau\}$ is a normal subgroup of $\{1,\sigma^2,\tau,\sigma^2\tau\}$ but $H\not\vartriangleleft D_4$.

Example. The center $Z(G)$ of a group $G$
$$Z(G)=\{x\in G: \forall g\in G, xg=gx\}$$ is a normal subgroup of $G$.

Example. If $N\leq Z(G)$ then $N\vartriangleleft G$.

Example. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\vartriangleleft G$ as seen here.

Example. Let $\mathbb{R}^\ast=\mathbb{R}\setminus\{0\}$. Define $f: \mathrm{GL}(2,\mathbb{R})\longrightarrow\mathbb{R}^\ast$ by
$$f(A)=\det A,\ \mbox{for any}\ A\in\mathrm{GL}(2,\mathbb{R}).$$ Then $f$ is a homomorphism because for any $A,B\in\mathrm{GL}(2,\mathbb{R})$, $f(AB)=\det(AB)=(\det A)(\det B)=f(A)f(B)$. $f$ is also onto: for any $r\in\mathbb{R}^\ast$ let $A=\begin{pmatrix} r & 0\\ 0 & 1 \end{pmatrix}$. Then $A\in\mathrm{GL}(2,\mathbb{R})$ and $f(A)=r$. For any $A\in\mathrm{GL}(2,\mathbb{R})$, $A\in\ker f$ if and only if $\det A=1$ and so $\ker f=\mathrm{SL}(2,\mathbb{R})$. Hence, $\mathrm{SL}(2,\mathbb{R})\vartriangleleft\mathrm{GL}(2,\mathbb{R})$.