Method 1:

Let $H$ be a subgroup of group $G$. Then $H \lhd G \iff \forall g \in G, \ gH = Hg \ (\text{or} \ gHg^{-1} = H)$.

Method 2:

Let $H$ be a subgroup of group $G$. Then $H \lhd G \iff \forall g \in G, h \in H, \ ghg^{-1} \in H$.

Definition 1. Let $H$ be a subgroup of a group $G$, and suppose that the index of $H$ in $G$ is $r$, that is, $[G:H]=r$. Then
$$
G=a_0H\cup a_1H\cup\cdots\cup a_{r-1}H,
$$
where $a_0=e$ and
$$
a_iH\cap a_jH=\varnothing \quad \text{for } i\ne j,
$$
is called a coset decomposition of $G$ with respect to its subgroup $H$.

Proposition 1

Let $G$ be a group, let $a\in G$, and suppose that the order of $a$ is $m$. Prove that:

(1) $a^n=e$ if and only if $m\mid n$.

(2) $a^k=a^h$ if and only if $k\equiv h \pmod m$.

Criterion 1: The Definition of a Group

Let $G$ be a nonempty set, and suppose a binary operation “$\cdot$” is defined on $G$. If this binary operation satisfies the following conditions:

(1) Closure: $ \forall x,y \in G,\ x\cdot y\in G $

(2) Associativity: $ \forall x,y,z\in G,\ (x\cdot y)\cdot z=x\cdot (y\cdot z) $

(3) Identity element: $ \exists e\in G,\ \forall a\in G,\ e\cdot a=a\cdot e=a $

(4) Inverse element: $ \forall a\in G,\ \exists b\in G,\ a\cdot b=b\cdot a=e $

then $G$ is called a group with respect to “$\cdot$”.