So far I have done the RHS to $(x-3)(y-3)+12$ and I have done this to find the identity element, $(e-3)(x-3)+12=x$, then re-arranged to $(e-3)=(x-12)/(x-3)$, now I am stuck.
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Apply definition of identity element. Put $y=0$ in $x*y=y$. – Kavi Rama Murthy Mar 20 '19 at 10:18
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1Again one of the various variations of this question. – Dietrich Burde Mar 20 '19 at 10:21
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$e=4$ will do the job for the identity. – little o Mar 20 '19 at 10:22
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You factored wrong. It factors as $(x-3)(y-3)+3$. – jgon Mar 20 '19 at 15:49
4 Answers
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The identity element (if it exists) is a value $e$ such that
$x*e = x \space \forall x$
Note that the symmetry of the definition of $x*y$ tells us that
$x*e = x \space \forall x \Rightarrow e*y = y \space \forall y$
so if we can find $e$ then it will be a two-sided identity.
$x*e = x \space \forall x \\ \Rightarrow xe -3x - 3e + 12 = x \space \forall x \\ \Rightarrow x(e-4) -3e +12 = 0 \space \forall x \\ \Rightarrow (x-3)(e-4)= 0 \space \forall x$
From this can you see a value for $e$ which will act as an identity ?
gandalf61
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Note that in any group $G$, the identity is the unique solution to the equation $g^2=g$.
Hence, the identity in your group is the unique solution to the equation $x^2-6x+12=x$ in ${\mathbf R}\setminus \{3\}$.
tomasz
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Hint: Show that the function $f(x)=x-3$ defines an isomorphism between your group and the group of nonzero real numbers with multiplication. Note that this way, you can simultaneously check that this formula really defines a group operation.
tomasz
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