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# problem about function and relation mappimg

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.

Note by Sayan Chaudhuri
3 years, 11 months ago

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may 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. · 3 years, 11 months ago