Please help.....

For the following sets, with the binary operation given, determine whether or not it forms a group, by checking the group axioms.

(R,◦),where x◦y=(xy)/√3

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- Nov 5th 2013, 04:43 AMFairmum2013Group axioms
Please help.....

For the following sets, with the binary operation given, determine whether or not it forms a group, by checking the group axioms.

(R,◦),where x◦y=(xy)/√3 - Nov 5th 2013, 05:37 AMSlipEternalRe: Group axioms
What are the group axioms? First, you need an identity element: $\displaystyle e\circ x = x\circ e = \dfrac{ex}{\sqrt{3}} = x$. By cancellation, we find $\displaystyle e = \sqrt{3}$. Next, check that associativity holds:

$\displaystyle \begin{align*}x\circ (y\circ z) & = \dfrac{ x \tfrac{yz}{ \sqrt{3} } }{ \sqrt{3} } \\ & = \dfrac{ \tfrac{xy}{\sqrt{3}} z }{ \sqrt{3} } \\ & = (x\circ y)\circ z\end{align*}$

This follows from associativity and commutativity of real numbers.

Next, check that each element has an inverse. Let $\displaystyle x \in \mathbb{R}$. You want to find $\displaystyle y \in \mathbb{R}$ so that $\displaystyle x\circ y = y\circ x = e = \sqrt{3}$. So, $\displaystyle x\circ y = \dfrac{xy}{\sqrt{3}} = \sqrt{3}$. Solving for $\displaystyle y,$ we find $\displaystyle y = \dfrac{3}{x}$. Hence, $\displaystyle y$ exists so long as $\displaystyle x \neq 0$.

So, no, it is not a group, but $\displaystyle \left(\mathbb{R}\setminus \{0\},\circ \right)$ is a group. - Nov 5th 2013, 02:02 PMFairmum2013Re: Group axioms
Much appreciated!!!!! I couldn't figure out the inverse.