Showing H is a normal subgroup by calculating left and right coset If $G = \begin{pmatrix} a & b \\
0 & 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$) 
and assume G is a group under matrix multpication
Prove that H = ($\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}: t\in\mathbb{R}$) is a normal subgroup of G
I'm meant to show that H is a normal subgroup by showing that the left coset is equal to the right coset.
So far I have shown the following:
take $g=\begin{pmatrix} a & b \\
0 & 1 \end{pmatrix}$ 
then 
$gH=\begin {pmatrix}a &at+b \\ 0 &1 \end{pmatrix}$
and take $g'=\begin{pmatrix} a' & b' \\
0 & 1 \end{pmatrix}$
then $Hg'=\begin{pmatrix} a' & b'+t \\
0 & 1 \end{pmatrix}$
I don't quite understand how $gH=Hg'$

First let me solve the problem the way you asked: proving that every left coset is a right coset.

Given $g=\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix}$ one must prove that $gH=Hg$ (introducing an additional symbol $g'$ is wrong). Your question correctly describes the general format of matrices in the sets $gH$ and $Hg$, so I'll work with that.
Consider $M$ in $gH$, and so there exist $t$ such that
$$M = \begin{pmatrix}a & at+b \\ 0 & 1 \end{pmatrix} \in gH
$$
For $M$ to be in $H g$, one must find $t'$ so that
$$M = \begin{pmatrix}a & b+t' \\ 0 & 1 \end{pmatrix} \in Hg
$$
Three of the matrix entries are already equal, so we must simply find $t'$ so that 
$$at+b = b+t'
$$
That equation is easily solved: $t'=at$, which proves that $M \in Hg$.
The reverse inclusion $Hg \subset gH$ is proved similarly.

However, despite the definition of a normal subgroup $H < G$ being 


*

*"every left coset of $H$ is a right coset of $H$"


and despite the utility of that definition in applying normality, for purposes of proving normality it is almost always easier  to work from the equivalent statement 


*

*"$gHg^{-1}=H$ for every $g \in G$"


or even easier


*

*"$g H g^{-1} \subset H$ for every $g \in G$"


So, given $g = \begin{pmatrix}a & b \\ 0 & 1 \end{pmatrix}$ and $h = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$, I simply do a calculation
$$g h g^{-1} = \begin{pmatrix}a & b \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\begin{pmatrix}a^{-1} & -b/a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & at+b \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a^{-1} & -b/a \\ 0 & 1\end{pmatrix} = \begin{pmatrix} 1 & at \\ 0 & 1\end{pmatrix}
$$
And this is obviously in $H$, by setting $t'=at$.
