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Math Help - A question regarding A(S) from baby Herstein

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

    Quote Originally Posted by Nikita2011 View Post
    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.
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    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
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    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.
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