# Thread: A question regarding A(S) from baby Herstein

1. ## A question regarding A(S) from baby Herstein

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 $F: S_m \to S_n$ such that $F(fg)=F(f)F(g)$ for all $f,g \in S_m$.

well, my idea is that with every element f in $S_m$ I associate an element f' in $S_n$ defined in this way:
If $n \in S_m$ then $f'(n)=f(m)$, otherwise: $f'(n)=n$.

then I define $(fog)'$ as $f'og'$. It's easy to show that this operation is well defined. Now, I define the map $F(fg)=F(f)F(g)$ by $F(f)=f'$.
To see that this map is one-to-one, suppose $F(f)=f'=g'=F(g)$. from the definition of $f'$, It's easy to see that $f=g$.
Now, with the definition of the function F, It's clear that $F(fg)=(fg)'=f'g'=F(f)F(g)$ and we're done.

2. ## Re: A question regarding A(S) from baby Herstein

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 $F: S_m \to S_n$ such that $F(fg)=F(f)F(g)$ for all $f,g \in S_m$.

well, my idea is that with every element f in $S_m$ I associate an element f' in $S_n$ defined in this way:
If $\color{blue}n \in S_m$ then $\color{blue}f'(n)=f(m)$, otherwise: $\color{blue}f'(n)=n$.

then I define $(fog)'$ as $f'og'$. It's easy to show that this operation is well defined. Now, I define the map $F(fg)=F(f)F(g)$ by $F(f)=f'$.
To see that this map is one-to-one, suppose $F(f)=f'=g'=F(g)$. from the definition of $f'$, It's easy to see that $f=g$.
Now, with the definition of the function F, It's clear that $F(fg)=(fg)'=f'g'=F(f)F(g)$ and we're done.
I think that you have the right idea here, but the definition in blue is completely garbled. It should read something like:

If $k\leqslant m$ then $f'(k) = f(k)$, otherwise $f'(k)=k.$

3. ## Re: A question regarding A(S) from baby Herstein

Yea. first I defined it the way you did, then I thought 'what if our members weren't numbers? then $\leq$ would be a meaningless relation on them'. Is my definition totally wrong or it's only garbled? xD

4. ## Re: A question regarding A(S) from baby Herstein

i assume by A(S) you mean the set of all bijections on a set S.

it is usually understood that the domain of the elements of $S_m$ is the set {1,2,...,m}.

if not, you can easily create a bijection between {1,2,...,m} and whatever m letters you are permuting with $S_m$ and induce an ordering on this set from that bijection.

EXAMPLE: suppose we represent S3 as:

A-->A
B-->B
C-->C for the identity

A-->B
B-->A
C-->C for the transposition (1 2), and so on.

here, the set S is {A,B,C}. so we have a bijection (i'll call it k, whatever).

1<-->A
2<-->B
3<-->C (that is k(1) = A, k(2) = B, k(3) = C).

for m,n in S, define m < n if and only if $k^{-1}(m) < k^{-1}(n)$.

all of which is a rather long-winded way of saying, you may as well assume that the set that $S_m$ is acting on is (set-isomorphic to) {1,2,...,m}.

the "numbers" $S_m$ "permutes" are just "placeholders", in many practical applications these may be algebraic objects (such as subgroups), or hydrogen atoms (in certain symmetry group applications in chemistry), or seating arrangments at a table. so, in many cases, when one refers to " $S_m$", one is refering to a group of functions that acts "just like" the set of bijections on the set of natural numbers {1,2,...,m}.

the natural bijection between the set {1,2,..m} and some given set of m elements is so transparent as to be omitted from most discussions.