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Math Help - group homomorphism

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    group homomorphism

    Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

    I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...
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    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Icarus0 View Post
    Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

    I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...

    what do you mean by Sym(n)
    you want to prove that f is sgn what do you mean by sgn
    write what you want without using shortcuts so all can understand your question
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    Quote Originally Posted by Icarus0 View Post
    Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

    I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...
    well, since f is surjective, we have [S_n: \ker f]= 2 but we know (i hope you know!) that A_n is the unique subgroup of S_n with index 2. thus \ker f = A_n and you're done.
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    Quote Originally Posted by Icarus0 View Post
    Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?


    No, it isn't. If there exists such an onto homomorphism then the group has to have even order...and not all (finite) groups have even order, of course.

    Tonio

    I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...
    .
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    Quote Originally Posted by Amer View Post
    what do you mean by Sym(n)
    you want to prove that f is sgn what do you mean by sgn
    write what you want without using shortcuts so all can understand your question
    Sym(n) is the group that consists of all permutations of the first n natural numbers. sgn is the sign of the permutation.

    well, since is surjective, we have but we know (i hope you know!) that is the unique subgroup of with index 2. thus and you're done.
    I don't understand anything... S_n, A_n???


    No, it isn't. If there exists such an onto homomorphism then the group has to have even order...and not all (finite) groups have even order, of course.

    Tonio
    Has sym(n) even order?
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    Quote Originally Posted by Icarus0 View Post
    Sym(n) is the group that consists of all permutations of the first n natural numbers. sgn is the sign of the permutation.

    I don't understand anything... S_n, A_n???


    Has sym(n) even order?

    Yes it does, for any n\geq 2...and it's Sym_n=:S_n\,,\,\,A_n= all the even permutations in S_n. If you don't know this then it will probably be a good idea you first read a little about this before you attempt to solve your question.

    Tonio
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    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Icarus0 View Post
    Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

    I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...

    s_n is the group of permutations A_n is the even permutation

    f : S_n \rightarrow \text{ {-1,1} }

    f is multiplicative and surjective want f is sgn

    f is multiplicative f(a.b) = f(a) . f(b)

    ker(f) = \left\{ \alpha \in S_n : f(\alpha) = identity\; of\; the\; range\; =1\right\}
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    Oh, I see. A_n is the kernel of sgn. We called this Alt(n).
    But still I don't understand the following
    well, since is surjective, we have but we know (i hope you know!) that is the unique subgroup of with index 2. thus and you're done.
    What is the index of a group and what does [S_n:ker f] mean?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Icarus0 View Post
    Oh, I see. A_n is the kernel of sgn. We called this Alt(n).
    But still I don't understand the followingWhat is the index of a group and what does [S_n:ker f] mean?
    I assume, and let NCA validated this, but since f:S_n\mapsto\mathbb{Z}_2 is surjective we have that S_n/\text{ker }f\cong\mathbb{Z}_2 and in particular that \left|S_n/\text{ker }f\right|=\left[S_n:\text{ker }f\right]=2 where \left[S_n:\text{ker }f\right] is the number of cosets formed by \text{ker }f (the index of \text{ker }f in {S}_n) therefore since A_n is the only subgroup of order \frac{n!}{2} it follows that \text{ker }f=A_n for every surjection. But, this forces f\left(S_n-A_n\right)=\{1\} which shows that it is the only surjection.
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