a function g from a set X to itself satisfies g^m=g^n for positive integers m and n with m>n.Here g^n stands for gogogo.......og(n times).Show that g is one one function iff g is onto function.

\(g^{(m)}=g^{(n)}\Rightarrow{g^{(m-n)}(x)=x}\) and for some \(x_{0}\in{Χ}\)

\(g^{(m-n)}(x_{0})=x_{0}\)

if we set \(g^{(m-n-1)}(x_{0})=a\) then
\(g(a)=x_{0}\) which implies that for every \(x_{0}\in{X},\exists{a}: g(a)=x_{0}\), which is the definition of a surjective function.

2)If g is onto,
and we set \(g^{(n)}(x_{0})=c\) for some \(x_{0}\) then, for all c, we have that \(g^{(m-n)}(c)=c\).What we don't know yet is whether an \(x_{0}\) can be assigned to every value of c.
But we can easily prove that if g is onto, \(g^{(n)}\) is onto. Now we can conclude that

\(g(x_{1})=g(x_{2})\Rightarrow{g^{(m-n)}(x_{1})=g^{(m-n)}(x_{2})}\Rightarrow{x_{1}=x_{2}}\)
which is valid for every \(x_{1},x_{2}\in{X}\) as we proved above, so g is 1-1.

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## Comments

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TopNewestmay I present a solution?

1)If g is one-to-one then:

\(g^{(m)}=g^{(n)}\Rightarrow{g^{(m-n)}(x)=x}\) and for some \(x_{0}\in{Χ}\)

\(g^{(m-n)}(x_{0})=x_{0}\)

if we set \(g^{(m-n-1)}(x_{0})=a\) then \(g(a)=x_{0}\) which implies that for every \(x_{0}\in{X},\exists{a}: g(a)=x_{0}\), which is the definition of a surjective function.

2)If g is onto, and we set \(g^{(n)}(x_{0})=c\) for some \(x_{0}\) then, for all c, we have that \(g^{(m-n)}(c)=c\).What we don't know yet is whether an \(x_{0}\) can be assigned to every value of c. But we can easily prove that if g is onto, \(g^{(n)}\) is onto. Now we can conclude that

\(g(x_{1})=g(x_{2})\Rightarrow{g^{(m-n)}(x_{1})=g^{(m-n)}(x_{2})}\Rightarrow{x_{1}=x_{2}}\) which is valid for every \(x_{1},x_{2}\in{X}\) as we proved above, so g is 1-1.

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which book is this excerpt from?

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