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## help with isomorphism

Groups, Rings, Domains, Modules, etc, Galois theory
tredy
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### help with isomorphism

I have a hard time to show the following isomorphism anyone can help me with the type $f : G\to {\mathbb{Q}}^{*}$ .

Show that the group $G={\mathbb{Q}}\setminus \{−1/2\}$ with binary operation $x\ast y=x+y+2xy$ is isomorphic to ${\mathbb{Q}}\setminus \{0\}$ with multiplication.
Also find a subgroup $H\leq G$, such that $G/H\cong {\mathbb{Z}}_2$ (the group with two elements)
Papapetros Vaggelis
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### Re: help with isomorphism

Hi Tredy.

Firstly, if $\displaystyle{x\,,y\in G}$, then

\displaystyle{\begin{aligned}x\star y+\dfrac{1}{2}&=x+y+2\,x\,y+\dfrac{1}{2}\\&= \dfrac{2\,x+1}{2}+\dfrac{2\,y\,(2\,x+1)}{2}\\&=\dfrac{(2\,y+1)\,(2\,x+1)}{2}\\&\neq 0\end{aligned}}

so, $\displaystyle{x\star y\neq -\dfrac{1}{2}\implies x\star y\in G}$.

Let $\displaystyle{x\,,y\,,z\in G}$. Then,

\displaystyle{\begin{aligned}(x\star y)\star z&=(x+y+2\,x\,y)\star z\\&=x+y+2\,x\,y+z+2\,(x+y+2\,x\,y)\,z\\&=x+y+2\,x\,y+z+2\,x\,z+2\,y\,z+4\,x\,y\,z \end{aligned}}

and

\displaystyle{\begin{aligned}x\star (y\star z)&=x\star (y+z+2\,y\,z)\\&=x+y+z+2\,y\,z+2\,x\,(y+z+2\,y\,z)\\&=x+y+z+2\,y\,z+2\,x\,y+2\,x\,z+4\,x\,y\,z\\&=(x\star y)\star z \end{aligned}}

Also, $\displaystyle{0\in G}$ and $\displaystyle{x\star 0=x+0+2\,x\,0=x=0\star x}$.

Obviously, the operation $\displaystyle{\star}$ is abelian at $\displaystyle{G}$.

Finally, let $\displaystyle{x\in G}$. Then, $\displaystyle{y=\dfrac{-x}{2\,x+1}\in G}$ and

$\displaystyle{2\,x\,y+y=-x\iff x+y+2\,x\,y=0\iff x\star y=0=y\star x}$, so

$\displaystyle{x^{-1}=-\dfrac{x}{2\,x+1}}$.

Therefore, $\displaystyle{\left(G,\star\right)}$ is an abelian group.

Now, define $\displaystyle{f:G=\mathbb{Q}\setminus \left\{-1/2\right\}\to \mathbb{Q}^{\star}}$

by $\displaystyle{f(x)=\dfrac{1}{2\,x+1}}$. Obviously, $\displaystyle{f}$ is $\displaystyle{1-1}$.

If $\displaystyle{x\,,y\in G}$, then,

$\displaystyle{f(x\star y)=f(x+y+2\,x\,y)=\dfrac{1}{2\,(x+y+2\,x\,y)+1}}$ and

\displaystyle{\begin{aligned} f(x)\,f(y)&=\dfrac{1}{2\,x+1}\,\dfrac{1}{2\,y+1}\\&=\dfrac{1}{4\,x\,y+2\,x+2\,y+1}\\&=\dfrac{1}{2\,(x+y+2\,x\,y)+1}\\&=f(x\star y)\end{aligned}}.

Therefore, the function $\displaystyle{f}$ is a group monomorphism.

Finally, let $\displaystyle{y\in\mathbb{Q}^{\star}}$. Then,

\displaystyle{\begin{aligned}f(x)=y&\iff \dfrac{1}{2\,x+1}=y\\&\iff 2\,x+1=\dfrac{1}{y}\\&\iff 2\,x=\dfrac{1-y}{y}\\&\iff x=\dfrac{1-y}{2\,y} \end{aligned}}

where $\displaystyle{x=\dfrac{1-y}{2\,y}\in G}$. Indeed,

$\displaystyle{x+\dfrac{1}{2}=\dfrac{1-y}{2\,y}+\dfrac{y}{2\,y}=\dfrac{1}{2\,y}\neq 0}$

So, $\displaystyle{f}$ is an isomorphism and $\displaystyle{\left(G,\star\right)\cong \left(\mathbb{Q}^{\star},\cdot\right)}$.

Let $\displaystyle{H=\mathbb{Q}\cap \left(-\dfrac{1}{2},+\infty\right)\subseteq G}$. We have

that $\displaystyle{e_{G}=0\in H}$. Also, if $\displaystyle{x\,,y\in H}$, then,

\displaystyle{\begin{aligned} x\star y^{-1}&=x\star \left(-\dfrac{y}{2\,y+1}\right)\\&=x-\dfrac{y}{2\,y+1}+2\,x\,\dfrac{-y}{2\,y+1}\\&=\dfrac{2\,x\,y+x-y-2\,x\,y}{2\,y+1}\\&=\dfrac{x-y}{2\,y+1}>-\dfrac{1}{2}\\&\iff \dfrac{x-y}{2\,y+1}+\dfrac{1}{2}>0\\&\iff \dfrac{2\,x-2\,y+2\,y+1}{2\,(2\,y+1)}>0\\&\iff \dfrac{2\,x+1}{2\,(2\,y+1)}>0\end{aligned}}

and the last one is true.

So, $\displaystyle{H\leq G}$.

Define $\displaystyle{g:G\to \mathbb{Z}_{2}\,,g(x)=0\,,x>-\dfrac{1}{2}\,\,\,,g(x)=1\,,x<-\dfrac{1}{2}}$.

If $\displaystyle{x\,,y\in G}$, then,

$\displaystyle{x\star y+\dfrac{1}{2}=\dfrac{(2\,y+1)\,(2\,x+1)}{2}}$ and

$\displaystyle{x\in H\,\,\land\,\,y\in H\implies g(x\star y)=0=g(x)+g(y)}$

$\displaystyle{x\in H\,\,\land\,\, y\notin H\implies g(x\star y)=1=g(x)+g(y)}$

$\displaystyle{x\notin H\,\,\land\,\, y\notin H\implies g(x\star y)=0=1+1=g(x)+g(y)}$.

In any case, $\displaystyle{g(x\star y)=g(x)+g(y)}$, so $\displaystyle{g}$ : homomorphism

which is onto $\displaystyle{\mathbb{Z}_{2}}$. Also,

$\displaystyle{\rm{Ker}(g)=\left\{x\in G\,,g(x)=0\right\}=H}$ and according to the 1st Isomorphism

Theorem, $\displaystyle{\left(G/H,\star\right)\cong \left(\mathbb{Z}_{2},+\right)}$.
tredy
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### Re: help with isomorphism

really impressive...well done.Respect