Let $x*y=xy-x-y+2$ for $x,y\in H=(1,\infty)$. I need to show that $(H,*)$ is isomorphic to $(\mathbb{R},+)$ as groups. I have already proven to myself that $H$ is a group, the identity is 2, the operation is associative, and that each element in $H$ has an inverse. But I cannot figure out how to construct a homomorphism let alone an isomorphism between the two groups. How would I prove this? Could it be done using a contradiction proof; like if I assumed they are not isomorphic and reached a contradiction? Any help is appreciated!

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    $\begingroup$ Hint: first try to show that $(H,\star)$ is isomorphic to $(\mathbb{R}^+, \times)$; there's then a well-known isomorphism between $(\mathbb{R}^+, \times)$ and $(\mathbb{R}, +)$ that you can compose with your first isomorphism. $\endgroup$ – Steven Stadnicki Feb 3 '17 at 2:06
  • $\begingroup$ @StevenStadnicki That makes sense, but how exactly would I show that $(H,*)\simeq (\mathbb{R}^{+},\cdot)$? $\endgroup$ – Sir_Math_Cat Feb 3 '17 at 2:08
  • $\begingroup$ The fact that $x*y=(x-1)(y-1)+1$ is likely useful here. $\endgroup$ – Paul Feb 3 '17 at 2:11
  • $\begingroup$ HInt: $xy-x-y+2 = (x-1)(y-1)+1$ so $(x*y) - 1 = (x-1)(y-1)$ $\endgroup$ – sas Feb 3 '17 at 2:11

$xy - x - y + 2 = (x-1)(y-1)+1$

$f(x) = x+1$

$f(xy) = f(x)*f(y) = xy+1$

$g(x) = e^x + 1$

$g(x+y) = e^{x+y} + 1 = g(x)*g(y) = (e^x + 1)*(e^y+1) = e^xy + 1$

does $g(x)$ have an inverse?

$g^{-1} = \ln (x -1)$


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