Show that $(G, +, 0)$ and $(H, +, 0_{2×2})$ are abelian groups. Let $G = \big\{a + b\sqrt2 | a,b \in\mathbb{Q}\big\}$.
Let $H = \bigg\{\begin{bmatrix} a & 2b \\ b & a \end{bmatrix}\bigg |a,b \in\mathbb{Q}\bigg\} $
And denote $0_{2\times 2} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$,
then I have to show that $(G, +, 0)$ and $(H, +, 0_{2×2})$ are abelian groups. I know that a group is abelian if $\forall x,y \in G$ we have $x * y = y * x$.
Now, my problem is that I am not quite sure how to construct this proof. So any help/tip/example would be grateful.
Thanks in advance.
 A: Here are some steps for constructing a proof that the group $G$ is abelian:

*

*Consider any two elements of $G$. These elements can be written in the form $g_1 = a_1 + b_1 \sqrt{2},$ $g_2 = a_2 + b_2 \sqrt{2}$.

*In terms of our variables $a_i$ and $b_i$, write $g_1 + g_2$ in the form $a + b \sqrt{2}$ for suitable $a,b \in \Bbb Q$

*Similarly, write $g_2 + g_1$ in the form $a + b \sqrt{2}$ for suitable $a,b \in \Bbb Q$

*Look at the results from the previous two steps. How can we conclude that $g_1 + g_2$ and $g_2 + g_1$ are equal? (What does it mean for two elements of $G$ to be equal, by the way?)

The proof for $H$ is essentially the same.
A: Actually, both groups are isomorphic: $G\cong H$, see
How to prove that two groups $G$ and $H$ are isomorphic?
So it suffices to show that, say, $H$ is abelian. But this is clear from
$$
\begin{pmatrix} a & 2b \cr b & a \end{pmatrix}
\begin{pmatrix} c & 2d \cr d & c \end{pmatrix}=
\begin{pmatrix} ac+2bd & 2(ad+bc) \cr ad+bc & ac+2bd \end{pmatrix}=
\begin{pmatrix} c & 2d \cr d & c \end{pmatrix}
\begin{pmatrix} a & 2b \cr b & a \end{pmatrix}
$$
