# Proving a homomorphism is surjective.

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!))

Thanks!

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 $$(G,*)$$?

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