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

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem.

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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.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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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.
 
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