I have to prove that $$3\mathbb{Z}/6\mathbb{Z} \trianglelefteq \mathbb{Z}/6\mathbb{Z}.$$
I proved $3\mathbb{Z}/6\mathbb{Z} \leq \mathbb{Z}/6\mathbb{Z}$, so I have to show
$\forall A \in \mathbb{Z}/6\mathbb{Z},$ $A^{-1}+3\mathbb{Z}/6\mathbb{Z}+A \subset 3\mathbb{Z}/6\mathbb{Z}$.
(Proof)
Let $A \in \mathbb{Z}/6\mathbb{Z}$.
I can write $A=a+6\mathbb{Z}$ where $a \in \mathbb{Z}$ and then $A^{-1}=-a+6\mathbb{Z}$.
I'm stuck here.
What I have to do next is letting $B \in A^{-1}+3\mathbb{Z}/6\mathbb{Z}+A $ and showing $B \in 3\mathbb{Z}/6\mathbb{Z}.$
But when I let $B \in A^{-1}+3\mathbb{Z}/6\mathbb{Z}+A $, how can I express $B$ ?
Since $G:=\Bbb Z/6\Bbb Z$ is abelian, each of its subgroups is normal. (See @azif00's excellent comment.)
Every quotient of an abelian group is abelian. But $\Bbb Z$ is abelian, so $6\Bbb Z$ is normal in $\Bbb Z$, so $\Bbb Z/6\Bbb Z$ is abelian.
It suffices, then, to show that $H:=3\Bbb Z/6\Bbb Z\le G$.
I will use the one-step subgroup test
Since $0\times 3=0\times 6=0$, we have $0+6\Bbb Z\in H$. Hence $H\neq\varnothing$.
Suppose $a+6\Bbb Z\in H$. Then $a\in 3\Bbb Z$. But $3\Bbb Z\subseteq \Bbb Z$. Hence $a+6\Bbb Z\in G$. Hence $H\subseteq G$.
Suppose $x=a+6\Bbb Z, y=b+6\Bbb Z\in H$, for some $a,b\in 3\Bbb Z$. Then
$$\begin{align} x-y&= (a+6\Bbb Z)-(b+6\Bbb Z)\\ &=(a+6\Bbb Z)+((-b)+6\Bbb Z)\\ &=(a+(-b))+6\Bbb Z\\ &=(a-b)+6\Bbb Z, \end{align}$$
but $a-b\in 3\Bbb Z$ since $3\Bbb Z\le \Bbb Z$. Hence $x-y\in H$.
Hence $H\le G$.
Hence $H\unlhd G$.