Originally Posted by

**Nikita2011** I just wanna know if my argument could be written down on my final exam paper as a correct formal argument.

This is the problem: If m<n, show that there is a 1-1 mapping $\displaystyle F: S_m \to S_n$ such that $\displaystyle F(fg)=F(f)F(g)$ for all $\displaystyle f,g \in S_m$.

well, my idea is that with every element f in $\displaystyle S_m$ I associate an element f' in $\displaystyle S_n$ defined in this way:

If $\displaystyle \color{blue}n \in S_m$ then $\displaystyle \color{blue}f'(n)=f(m)$, otherwise: $\displaystyle \color{blue}f'(n)=n$.

then I define $\displaystyle (fog)'$ as $\displaystyle f'og'$. It's easy to show that this operation is well defined. Now, I define the map $\displaystyle F(fg)=F(f)F(g)$ by $\displaystyle F(f)=f'$.

To see that this map is one-to-one, suppose $\displaystyle F(f)=f'=g'=F(g)$. from the definition of $\displaystyle f'$, It's easy to see that $\displaystyle f=g$.

Now, with the definition of the function F, It's clear that $\displaystyle F(fg)=(fg)'=f'g'=F(f)F(g)$ and we're done.