Let G be a group, for all g∈G we define the function: φ_g:G→G as φ_g(x)=gxg^(-1)
Show that the function is surjective and injective.
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Particularly
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Do you know what a "group" is? Do you know what "surjective" or "injective" mean? If you expect to be able to answer this problem, you certainly should but you give no indication that you do.
Even if you do not know how to start this problem stating those definitions will help us point out how to use them.
I apologyze, I have used the inverse of the elements in the group in order to show that when I have two elements that are evaluated by the function, they will be equal if and only if, the elements are also equal. The other part, for a function in order to be surjective, I need to show that for all y in G there exists an x in G such that when x is evaluated by the function, it will equal y. But I am not sure if this second part may be shown by the inverse of the function. That is where I am having trouble mostly.
Thanks for the recommendations.
We need, given any $y \in G$, to find an $x \in G$ such that $\phi_g(x) = y$.
Concretely, this means we need to find an $x$ so that: $gxg^{-1} = y$.
This is an equation involving a product in a group. We can solve it. What might you multiply both sides by, to get just $x$, on the left side of the equation?