So what I did was say that for every $x$ in $\gamma$ we have $\alpha(\gamma(x))=\beta(\gamma(x))$ and since $\gamma$ is bijective $x$ gets mapped to only one element of the range and will call this $y$. Now we have $\alpha(y)=\beta(y)$ and then i basically said since $\alpha$ and $\beta$ are both functions and there range is equal the source from the domain can't be different by the definition of a function. But after thinking for awhile I decided that I am probably wrong. The hint i received was to apply $\gamma^{-1}$. any help would be appreciated.
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Hint: $(\alpha\circ\gamma)\circ\gamma^{-1}=(\beta\circ\gamma)\circ\gamma^{-1}$
egreg
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That's what the hint was saying. If I do that I would write it like this $\alpha(\gamma(\gamma^{-1})) = \beta(\gamma(\gamma^{-1}))$ and then would the $\gamma$'s just cancel out? leaving $\alpha=\beta$? – L. Johnson Sep 02 '16 at 13:07
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yes, because $\gamma$ is bijective and composition of functions is associative – Sep 02 '16 at 13:08
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for some reason i wasn't thinking about the composition of functions being associative. Thank you – L. Johnson Sep 02 '16 at 13:10