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For group $(\Bbb R\setminus\{−10/3\},\ast)$, where the operation $\ast$ is given by: $$x\ast y = 3xy+ 10x+ 10y+ 30.$$

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  1. Group has identity

  2. each member in group has inverse

  3. is group abelian?

The problem also states that it is closed under $\ast$ and is associative

I think I understand the properties of each, but don't know how to go about solving. Thanks in advance

Shaun
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mathguy1234
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2 Answers2

1

The key observation is that $\ast$ is just ordinary multiplication in disguise.

The function $f(x)=3x+10$ defines a bijection $\mathbb R\setminus\{−10/3\} \to \mathbb R\setminus\{0\}$. Moreover, $x \ast y= f^{-1}(f(x)f(y))$. Therefore, the properties of $\ast$ correspond exactly to the properties of ordinary multiplication. In particular, $\ast$ defines an abelian group isomorphic to the multiplicative group $\mathbb R^*$. The group identity is $f^{-1}(1)=-3$.

This technique is called transport of structure.

lhf
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0

For (1), since the only idempotent of a group is the identity$^\dagger$, solve $x\ast x=x$ for $x$.

$$3x^2+20x+30=x$$ has solutions $$x=\frac{-19\pm\sqrt{19^2-4(3)(30)}}{6},$$ i.e., $$x=-3\quad\text{or}\quad \color{red}{x=-10/3}.$$

For (2), plug your solution $e$ to (1) in $x\ast y=e$; solve for $y$ in terms of $x$.

Fix $x\in\Bbb R\setminus \{-10/3\}$. Solve $$3xy+10x+10y+30=-3;$$ that is, $$y=\frac{-10x-33}{3x+10}.$$

For (3), since $3xy=3yx$ and $10x+10y=10y+10x$, the operation is abelian.

$\dagger$: Suppose otherwise, that there is an idempotent $y\in G$ such that $y\neq e$. Then $y=y^2$ by definition. Multiply on the left, say, by $y^{-1}$; then

$$\begin{align} e&=y^{-1}y\\ &=y^{-1}(yy)\\ &=(y^{-1}y)y\\ &=ey\\ &=y, \end{align}$$

a contradiction.

Shaun
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