\(\mbox{Let, }\varphi:G\rightarrow H\mbox{ such that, }\varphi:a+b\sqrt{2}\mapsto\begin{bmatrix} a & 2b \\ b & a\end{bmatrix}\mbox{ where }a,\,b\in\mathbb{Q}\)First we shall show that \(\varphi\) is a well defined, bijective function. Take any two elements; \(a_{1}+b_{1}\sqrt{2},\,a_{2}+b_{2}\sqrt{2}\in G\) such that, \(a_{1}+b_{1}\sqrt{2}=a_{2}+b_{2}\sqrt{2}\mbox{ where }a_{1},\,a_{2},\,b_{1},\,b_{2}\in\mathbb{Q}\,.\)\[\Rightarrow (a_{1}-a_{2})+(b_{1}-b_{2})\sqrt{2}=0\]Since, \(a_{1}-a_{2},\,b_{1}-b_{2}\in\mathbb{Q}\) it can be easily shown that,\[a_{1}-a_{2}=b_{1}-b_{2}=0\]\[\Rightarrow a_{1}=a_{2}\mbox{ and }b_{1}=b_{2}\]\[\therefore\begin{bmatrix} a_{1} & 2b_{1} \\ b_{1} & a_{1}\end{bmatrix}=\begin{bmatrix} a_{2} & 2b_{2} \\ b_{2} & a_{2}\end{bmatrix}\]\[\mbox{That is }\varphi\mbox{ is a well defined function.}~~~~~~~~~~(1)\]Take any two elements; \(a_{1}+b_{1}\sqrt{2},\,a_{2}+b_{2}\sqrt{2}\in G\) such that, \(\varphi(a_{1}+b_{1}\sqrt{2})=\varphi(a_{2}+b_{2}\sqrt{2})\,.\) Then,\[\begin{bmatrix} a_{1} & 2b_{1} \\ b_{1} & a_{1}\end{bmatrix}=\begin{bmatrix} a_{2} & 2b_{2} \\ b_{2} & a_{2}\end{bmatrix}\]\[\Rightarrow a_{1}=a_{2}\mbox{ and }b_{1}=b_{2}\]\[\mbox{That is }\varphi\mbox{ is injective.}~~~~~~~~~~(2)\]Take any \(\begin{bmatrix} a & 2b \\ b & a\end{bmatrix}\in H\,.\) Then there exist \(a+b\sqrt{2}\in G\) such that,\[\varphi(a+b\sqrt{2})=\begin{bmatrix} a & 2b \\ b & a\end{bmatrix}\] \[\mbox{Therefore }\varphi\mbox{ is surjective.}~~~~~~~~~~(3)\]Take any \(x=a_{1}+b_{1}\sqrt{2},\,y=a_{2}+b_{2}\sqrt{2}\in G\) and consider \(\varphi(x+y)\,.\)\begin{eqnarray}\varphi(x+y)&=&\varphi\left((a_{2}+a_{2})+(b_{1}+b_{2})\sqrt{2}\right)\\&=&\begin{bmatrix} a_{1}+a_{2} & 2(b_{1}+b_{2}) \\ b_{1}+b_{2} & a_{1}+a_{2}\end{bmatrix}\\&=&\begin{bmatrix} a_{1} & 2b_{1} \\ b_{1} & a_{1}\end{bmatrix}+\begin{bmatrix} a_{2} & 2b_{2} \\ b_{2} & a_{2}\end{bmatrix}\\&=&\varphi(x)+\varphi(y)\end{eqnarray}\[\therefore\varphi(x+y)=\varphi(x)+\varphi(y)\, \forall\,x,\,y\in G~~~~~~~~~~~~(4)\]By (1), (2), (3) and (4);\[(G,+)\cong(H,+)\]We shall show that \(G\) and \(H\) are isomorphic under multiplication with respect to the same function, \(\varphi\,.\) Consider, \(\varphi(x\cdot y)\) where \(x=a_{1}+b_{1}\sqrt{2},\,y=a_{2}+b_{2}\sqrt{2}\in G\)\begin{eqnarray}\varphi(xy)&=&\varphi\left((a_{1}+b_{1}\sqrt{2})(a_{2}+b_{2}\sqrt{2})\right)\\&=&\varphi\left((a_{1}a_{2}+2b_{1}b_{2})+(a_{1}b_{2}+b_{1}a_{2})\sqrt{2}\right)\\&=&\begin{bmatrix} a_{1}a_{2}+2b_{1}b_{2} & 2(a_{1}b_{2}+b_{1}a_{2}) \\ a_{1}b_{2}+b_{1}a_{2} & a_{1}a_{2}+2b_{1}b_{2}\end{bmatrix}\\&=&\begin{bmatrix} a_{1} & 2b_{1} \\ b_{1} & a_{1}\end{bmatrix}\begin{bmatrix} a_{2} & 2b_{2} \\ b_{2} & a_{2}\end{bmatrix}\\&=&\varphi(x).\varphi(y)\end{eqnarray}\[\therefore\varphi(x\cdot y)=\varphi(x)\cdot\varphi(y)\, \forall\,x,\,y\in G\]Since we have already shown that \(\varphi\) is a bijective function,\[\therefore (G,\,\cdot)\cong(H,\,\cdot)\]Q.E.D.
