# Problem of the Week #11 - June 11th, 2012

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem.

-----

Problem: Let $G=\{a+b\sqrt{2} : a,b\in\mathbb{Q}\}$ and let $\displaystyle H=\left\{\begin{bmatrix} a & 2b \\ b & a\end{bmatrix} : a,b\in\mathbb{Q}\right\}$ be two groups. Show that $G$ and $H$ are isomorphic as groups under addition; i.e. find a bijective map $\varphi:G\rightarrow H$ such that for any $x,y\in G$, $\varphi(x+y) = \varphi(x) + \varphi(y)$, where $\varphi(x),\varphi(y)\in H$.

Are $G$ and $H$ isomorphic under multiplication? If yes, prove it. If not, provide a counterexample.

-----

Last edited:

#### Chris L T521

##### Well-known member
Staff member
This problem was correctly answered by Sudharaka. You can find his solution below.

$$\mbox{Let, }\varphi:G\rightarrow H\mbox{ such that, }\varphi:a+b\sqrt{2}\mapsto\begin{bmatrix} a & 2b \\ b & a\end{bmatrix}\mbox{ where }a,\,b\in\mathbb{Q}$$

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.

Status
Not open for further replies.