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
8 years, 4 months ago

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Comments

which book is this excerpt from?

Soham Chanda - 8 years, 4 months ago

may I present a solution?

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

g(m)=g(n)⇒g(m−n)(x)=xg^{(m)}=g^{(n)}\Rightarrow{g^{(m-n)}(x)=x} and for some x0∈Xx_{0}\in{Χ}

g(m−n)(x0)=x0g^{(m-n)}(x_{0})=x_{0}

if we set g(m−n−1)(x0)=ag^{(m-n-1)}(x_{0})=a then g(a)=x0g(a)=x_{0} which implies that for every x0∈X,∃a:g(a)=x0x_{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)(x0)=cg^{(n)}(x_{0})=c for some x0x_{0} then, for all c, we have that g(m−n)(c)=cg^{(m-n)}(c)=c.What we don't know yet is whether an x0x_{0} can be assigned to every value of c. But we can easily prove that if g is onto, g(n)g^{(n)} is onto. Now we can conclude that

g(x1)=g(x2)⇒g(m−n)(x1)=g(m−n)(x2)⇒x1=x2g(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 x1,x2∈Xx_{1},x_{2}\in{X} as we proved above, so g is 1-1.

Kyriakos Grammatikos - 8 years, 4 months ago
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