For reference, I'm linking the previous exercise - they're related.

This time, I want to prove that $ø$, the mapping $ø$: $(G,o)$ ---> $(G,*)$, defined by ø = $x_0^{-1}x$, for every $x\in G$.

So, I know that a function is surjective iff its inverse function domain is equal to the function's "arrival set" (don't know the proper English term - the codomain is contained into this larger set).

Let $y \in (G,*)$. We want $x \in (G,o)$ such that $ø(x)=y$; i.e., $x_0^{-1}x = y$.

I was a bit stuck, until I realised $x_0^{-1}$ is $1_{(G,\ast)}$ (check previous exercise). So this might help relating the two groups. Where do I go from here? How can I take advantage of having $1_{(G,\ast)}$ in the homomorphism's definition? Is there any place I can apply each group's operation?

((in the meantime, I'll be reading whether or not each group's operation affects a homomorphism!))


EDIT: Is it possible that all $ø$ does is take $x \in (G,o)$ and returns the very same element $x$, but this time in $(G,*)$?

  • $\begingroup$ Since the mapping $ø$ is from $(G,o)$ to $(G,*)$, then $ø(x) = x_0^{-1} \times x = x$, because $x_0^{-1}$ is identity element in $(G,*)$. So $ø$ is bijective, thus surjective. Is this right? $\endgroup$ – AJ44 Jan 22 '16 at 0:52

Let $\phi(g) = x_0^{-1}g$ and $a\ast b = a x_0 b$.

Let $y \in (G, \ast)$. Then $x_0 y \in (G, o)$ maps to $y$:

$$ \phi(x_0 y) = \phi(x_0) \ast \phi(y) = x_0^{-1} x_0 \ast x_0^{-1} y = e x_0 x_0^{-1} y = y$$

hence $\phi$ is surjective.

  • $\begingroup$ Thanks for the reply! But I need some clarification here, if you don't mind. Why pick $x_0 y$ as the element of $(G,o)$? $\endgroup$ – AJ44 Jan 22 '16 at 11:06
  • 1
    $\begingroup$ @AJ44 Because that is the element that is mapped to $y$ by $\phi$. $\endgroup$ – a student Jan 23 '16 at 1:37

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