Prove that $H$ is normal subgroup of $G$ I have a following question.
Let $p$ be a prime and let $G$ be a group and $H$ be a subgroup of $G$.
$$
G = \left\{
\begin{bmatrix}
a & b \\
0 & 1 
\end{bmatrix}
: a,b \in \mathbb{Z}_p, a \neq 0
\right\}$$
and 
$$ H = \left\{
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
: b \in \mathbb{Z}_p\right\}
$$
prove that $H$ is normal subgroup.
In my textbook, it gives two types of definition for normal subgroup.
The first one is if left coset and right coset are same with the subgroup, the subgroup is normal.
The other definition is let $H$ be a subgroup of $G$. Then $H$ is a normal subgroup of $G$ iff $xhx^{-1} \in H$ for every $h \in H$ and every $x \in G$.
I have tried with the first definition and got 
$$
xH = \left\{ \begin{bmatrix}
a & ac+b \\
0 & 1
\end{bmatrix} \right\}
$$
and for right coset
$$
Hx = \left\{ \begin{bmatrix}
a & b+c \\
0 & 1
\end{bmatrix} \right\}
$$ and I can't see that $xH \subset Hx$ and $Hx \subset xH$.
And I don't know how to apply the second definition.
Need help!

to follow Solid Snake's suggestion it will help if you know that
$$
\begin{bmatrix}
a & b \\
0 & 1 
\end{bmatrix}^{-1} = \begin{bmatrix}
a^{-1} &-ba^{-1}\\
0 & 1 
\end{bmatrix}
$$
