Let $N \subset G$ be a subgroup. Recall a definition.
A subgroup $N \subset G$ is normal if for any element $x \in N$ and any $g \in G$ the conjugate
$$
g x g^{-1}
$$
also belongs to $N$.
- If group $G$ is abelian, then any its subgroup is automatically normal, since $g x g^{-1}=x$.
- The alternating group $A_{n} \subset S_{n}$ is normal, since the conjugates
$$
\text { odd } \cdot \text { even } \cdot \text { odd }^{-1} \quad \text { even } \cdot \text { even } \cdot \text { even }^{-1}
$$
are always even.
- If $N$ is the kernel of a homomorphism $f: G \rightarrow H$, then $N$ is normal. Indeed, if $f(x)=e_{H}$, then
$$
f\left(g x g^{-1}\right)=f(g) f\left(g^{-1}\right)=e_{H},
$$
so $g x g^{-1} \in \operatorname{Ker}(f)$
- The subgroup $S L_{n}(\mathbb{R})$ of $\left(G L_{n}(\mathbb{R}), \cdot\right)$ is defined as the kernel of the homomorphism det : $\left(G L_{n}(\mathbb{R}), \cdot\right) \rightarrow\left(\mathbb{R}^{\times}, \cdot\right)$, , therefore it is normal.
The following proposition gives several equivalent convenient ways to state that $N \subset G$ is normal
Denote
$$
g N g^{-1}:=\left{g x g^{-1} \mid x \in N\right} \subset G
$$
The following properties of a subgroup $N \subset G$ are equivalent
- $N$ is normal;
- For any $g \in G$ we have
$$
g N g^{-1}=N
$$
- For any $g \in G$ the left and right cosets of $N$ generated by $g$ coincide:
$$
g N=N g
$$
Proof. We start by proving the implication $1 \Rightarrow 2$. If $N$ is normal, then we have that for every $g \in G$ $g N g^{-1} \subset N .$ It remains to prove the reverse inclusion. Let $x \in N$, and consider the element $k=g^{-1} x g .$ Since $N$ is normal, we have $k \in N$. Then $x=g k g^{-1}$ is an element of $g N g^{-1}$, so $N \subset g N g^{-1}$.
We now prove $2 \Rightarrow 3$. Let $g \in G$, and let $x \in N$. Then $g x=g x g^{-1} \cdot g \in N g$. Thus, we have $g N \subset N g$. In the same manner, we get $N g \subset g N$.
Finally we prove $3 \Rightarrow 1$. Let $g \in G$ and $x \in N$. We want to show that $g x g^{-1} \in N$, i.e., that $g x \in N g$. Since $g x \in g N$ which is equal to $N g$ by assumption, we are done.
If group $G$ has only one subgroup $H$ of size $r$, then $H$ is normal. Indeed, for any $g \in G$ the set $g H g^{-1}$ is also a subgroup of $G$. Since the map
$$
H \rightarrow g H g^{-1} \quad x \mapsto g x g^{-1}
$$
is a bijection (see homework # 9$), g H g^{-1}$ is also a subgroup of size $r .$ By assumption, such subgroup is unique, so
$$
g H g^{-1}=H,
$$
implying via part 2 of the proposition that $H$ is normal.
bgroup $H \subset G$ of index 2 is normal.
Proof. Indeed, if $H \subset G$ has index two, then the left/right cosets of $H$ are $H$ and $G \backslash H .$ Since the left and right cosets coincide, part 3 pf Proposition 1 implies that $H$ is normal.
It is useful to keep in mind also non-examples.
Subgroup $H={\mathrm{id},(12)} \subset S_{3}$ is not normal. Indeed, the left coset generated by (23) is
$$
(23) H={(23),(23)(12)}={(23),(132)},
$$
while the right coset is
$$
H(23)={(23),(12)(23)}={(23),(123)} .
$$