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

Question

question: If $m<n$, show that there is a $1$-$1$ mapping $F:S_m\rightarrow S_n$ such that $F(fg)=F(f)F(g)$ for all $f,g\in S_m$. Where $S_n$ stands for symmetric group of degree $n$

my approach:
First I was thinking how to construct the $F$. Because if I did this properly then the only work is left to show this is $1$-$1$ map.
So there are more elements in $S_n$ compare to $S_m$. I need to define $F$ in such a way that after covering all elements of $S_m$ I have to define it for other elements of $S_n$. Now here is my question arrive How to choose those elements and how to define $F$ when $S_m$ is covered?

Answer 1

You have the wrong approach to this question. You cannot guarantee with your construction that $F$ is a homomorphism. Here is the correct answer,

If $\sigma:\{1,...,m \} \rightarrow \{1,...,m \}$ is a bijection we define $F(\sigma):\{1,...,n\} \rightarrow \{1,...,n\}$ as follows

$$F(\sigma)(x) = x \ \text{ if } \ x > m$$ and

$$F(\sigma)(x) = \sigma(x) \ \text{ if } \ x \leq m$$

Intuitively $F \sigma$ just fixes the elements which are not in $\sigma$'s domain.

Try and prove that $F$ is a homomorphism.

Answer 2

Define $F:S_m\rightarrow S_n$ by $F(\sigma)(j)=\sigma(j)$ if $1\leq j\leq m$, and $F(\sigma)(j)=j$ if $m<j\leq n$.

The question just wants you to show that $S_n$ contains an isomorphic copy of $S_m$ when $n>m$.

Answer 3

More generally, take any set $Y$ and $X\subseteq Y$. The map $\phi \colon S_X\to S_Y$, defined by $\phi(\sigma)_{\mid X}:=\sigma$ and $\phi(\sigma)_{\mid Y\setminus X}:=Id_{Y\setminus X}$, is:

• "well defined": in fact, for every $\sigma\in S_X$, actually $\phi(\sigma)\in S_Y$;
• injective: in fact, $\phi(\sigma)=\phi(\tau)\Longrightarrow$ $\phi(\sigma)_{\mid X}=\phi(\tau)_{\mid X}\Longrightarrow$ $\sigma =\tau$;
• operation-preserving: in fact, for $x\in X$, $\phi(\sigma\tau)(x)=$ $(\sigma\tau)(x)=\sigma(\tau(x))=$ $\sigma(\phi(\tau)(x))\stackrel{\phi(\tau)(x)\in X}{=}$ $\phi(\sigma)(\phi(\tau)(x))=$ $(\phi(\sigma)\phi(\tau))(x)$; and, for $x\in Y\setminus X$, $\phi(\sigma\tau)(x)=$ $Id_{Y\setminus X}(x)\color{red}{=}$ $Id_{Y\setminus X}(Id_{Y\setminus X}(x))=$ $Id_{Y\setminus X}(\phi(\tau)(x))\stackrel{\phi(\tau)(x)\in Y\setminus X}{=}$ $\phi(\sigma)(\phi(\tau)(x))=$ $(\phi(\sigma)\phi(\tau))(x)$; therefore, $\phi(\sigma\tau)=\phi(\sigma)\phi(\tau)$.

Note that the equal sign in red holds only for the extension by the identity map of $Y\setminus X$; any other bijection on this set would fail the job.

So, $S_X\stackrel{\phi}{\hookrightarrow}S_Y$, and yours is just the particular case $Y=\{1,\dots,n\}$ and $X=\{1,\dots,m\}$.