Show that $H < GL(2,\Bbb R).$ Let
$$ H:= \left\{{{ \begin{bmatrix}a&b\\c&d\end{bmatrix} : a+b-c-d=0 }}\right\}. $$
Show that $H \leq \mathrm{GL}(2,\mathbb{R})$ where the symbol $\leq$ means subgroup.

It's easy to check that the identity matrix is in $H$ and so too that the inverse law holds but I'm struggling a bit with showing closure. I need to show:
$$ \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}e&f\\g&h\end{bmatrix}$$
is an element of $H$, but when I multiply it out
$$\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix}, $$
and put it into the "special" condition that $H$ has:
$$ (ae+bg)+(af+bh)-(ce+dg)-(cf+dh)=0, $$
I can't quite get the manipulation right as to get it into the form where the expression ends up equalling zero. I feel it is one of those, what I like to call, "flair moves" of adding and subtracting the same thing but I can't seem to quite see what to add/subtract in order to get it into a nice expression.
Any help would be much appreciated for this. Thanks in advance!

EDIT: Thank you all for your input! Ultimately, the fact that the matrix $$ \begin{bmatrix}e&f\\g&h\end{bmatrix},$$
is in H also, rather stupidly, flew right over my head which would've made the calculations a lot easier.
Thanks again for all the algebraic manipulations, including the "adding and subtracting the same thing" method and also, the problem in a bit more of a linear algebra concept involving vector spaces. It is all much appreciated.
 A: Let us rewrite the condition as $a+b=c+d=x$. It means that $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}x\\x\end{bmatrix},$$
meaning that the vector space $W$ of vectors having the same coordinates is an eigenspace for every matrix in $H$. But then, since the product of matrix is just the application of operators, if $A$ and $B$ are in $H$, then $AB$ will still have $W$ as an eigenspace, and so it is in $H$.
A: $(ae+bg)+(af+bh)−(ce+dg)−(cf+dh) = a(e+f) - c(e+f) + b(g+h) - d(g+h) = (a-c)(e+f) + (b-d)(g+h) =(a-c)(e+f) + (b-d)(e+f) = (a+b-c-d)(e+f) = 0 (e+f) = 0  $
A: You must remember that both of these matrices are in $H$.
So you have $a+b-c-d=0$ and $e+f-g-h=0$ to work with.
$$(ae+bg)+(af+bh)-(ce+dg)-(cf+dh)$$
$$=a(e+f)-c(e+f)+b(g+h)-d(g+h)= (a-c)(e+f)+(b-d)(g+h)$$
Using $g+h=e+f$ and then $a+b-c-d=0$ we obtain:$$=(a-c)(e+f)+(b-d)(e+f)=(a-c+b-d)(e+f)=0$$
A: The product of matrices in $H$ lies again in $H$, which you correctly translated into the condition:
$$
(ae+bg)+(af+bh)-(ce+dg)-(cf+dh)=0.
$$
Let's see why.
Since both matrices lie in $H$, you know that $a+b-c-d=0$ and $e+f-g-h=0$, which implies in particular that $e+f=g+h$. 
You want to show that
$$
ae+bg+af+bh-ce-dg-cf-dh=0,
$$
so you rewrite as
$$
(a-c)e+(a-c)f+(b-d)g+(b-d)h=0,
$$
so you want to show that
$$
(a-c)(e+f)+(b-d)(g+h)=0.
$$ 
If you substitute $g+h=e+f$, you get:
$$
(a-c+b-d)(e+f)=0,
$$
but the first factor is null, so you get the thesis.
A: \begin{eqnarray*}
 & & (ae+bg)+(af+bh)-(ce+dg)-(cf+dh)\\ &=& (a-c) (e+f)+(b-d)(g+h) \\
 &=& (a-c) (e+f)+\color{red}{(b-d) (e+f)-(b-d) (e+f)}+ (b-d)(g+h) \\
 &=& \underbrace{(a-c+b-d)}_{=0} (e+f)-(b-d) \underbrace{(e+f-g-h)}_{=0} \\
\end{eqnarray*}
