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Math Help - Signum function on symmetric groups

  1. #1
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    Signum function on symmetric groups

    Let \Omega be a set. Any g \in \text{Sym}(\Omega) can be written uniquely as a cycle decomposition in the form g = c_1c_2...c_k. Define \text{sgn }: \text{Sym}(\Omega) \rightarrow \{-1,1\} by \text{sgn}(g) = (-1)^{n-k}, where k is the number of cycles in the cycle decomposition of g and n is the size of \Omega.

    a). Let g,t \in \text{Sym}(\Omega) with t a transposition. Prove that \text{sgn}(tg) = -\text{sgn}(g).

    b). Show that for all g,h \in \text{Sym}(\Omega), \text{sgn}(gh) = \text{sgn}(g)\text{sgn}(h).

    These are both giving me a lot of trouble.. any ideas?
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  2. #2
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    Quote Originally Posted by dancavallaro View Post
    Let \Omega be a set. Any g \in \text{Sym}(\Omega) can be written uniquely as a cycle decomposition in the form g = c_1c_2...c_k. Define \text{sgn }: \text{Sym}(\Omega) \rightarrow \{-1,1\} by \text{sgn}(g) = (-1)^{n-k}, where k is the number of cycles in the cycle decomposition of g and n is the size of \Omega.

    a). Let g,t \in \text{Sym}(\Omega) with t a transposition. Prove that \text{sgn}(tg) = -\text{sgn}(g).

    b). Show that for all g,h \in \text{Sym}(\Omega), \text{sgn}(gh) = \text{sgn}(g)\text{sgn}(h).

    These are both giving me a lot of trouble.. any ideas?
    We can just think of \Omega = \{1,2,...,n\}.

    Without lose of generality say that t=(12). Let g be a non-identity element with cycle decomposition g=c_1c_2...c_k where these cycles are non-identity cycles. The way you prove this is by exploring various cases. The first case is that 1,2 are not found as entries in any one of the cycles and so tg has cycle decomposition (12)c_1...c_k and now apply the definition of the signum function. The second case is that 1,2 are found as entries in the cycle but 1,2 are found both in a single cycle c_1 (by relabeling if necessary). Therefore, we can think of c_1 as c=(1,a_1,...,a_j,2,b_1,...,b_i). Now (12)c_1 = (1,a_1,...,a_j)(2,b_1,...,b_i) and so (12)g = (1,a_1,...,a_j)(2,b_1,...,b_i)c_2...c_k and apply the definition of signum function. Another case is when 1,2 are found as entries in the cycle by 1,2 are found in different cycles say c_1,c_2 (by relabeling if necessary). So say that c_1 = (1,a_1,...,a_j) and c_2 = (2,b_1,...,b_k) now compute (12)c_1c_2 and apply the definition again. And finally there are cases when 1\text{ or }2 (but not both) appear as entries in one of the cycles. Just do all these cases and the proof ought to be complete. For the second part I imagine you need to use the result that every permutation is a product of transpositions.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    We can just think of \Omega = \{1,2,...,n\}.

    Without lose of generality say that t=(12). Let g be a non-identity element with cycle decomposition g=c_1c_2...c_k where these cycles are non-identity cycles. The way you prove this is by exploring various cases. The first case is that 1,2 are not found as entries in any one of the cycles and so tg has cycle decomposition (12)c_1...c_k and now apply the definition of the signum function. The second case is that 1,2 are found as entries in the cycle but 1,2 are found both in a single cycle c_1 (by relabeling if necessary). Therefore, we can think of c_1 as c=(1,a_1,...,a_j,2,b_1,...,b_i). Now (12)c_1 = (1,a_1,...,a_j)(2,b_1,...,b_i) and so (12)g = (1,a_1,...,a_j)(2,b_1,...,b_i)c_2...c_k and apply the definition of signum function. Another case is when 1,2 are found as entries in the cycle by 1,2 are found in different cycles say c_1,c_2 (by relabeling if necessary). So say that c_1 = (1,a_1,...,a_j) and c_2 = (2,b_1,...,b_k) now compute (12)c_1c_2 and apply the definition again. And finally there are cases when 1\text{ or }2 (but not both) appear as entries in one of the cycles. Just do all these cases and the proof ought to be complete. For the second part I imagine you need to use the result that every permutation is a product of transpositions.
    Thanks, that makes perfect sense! For part b), a hint I was given was to use the result in part a) as a base case for an inductive argument. But I can't really figure out what to induct on, or how to do it. Any suggestions?
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  4. #4
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    Quote Originally Posted by dancavallaro View Post
    Thanks, that makes perfect sense! For part b), a hint I was given was to use the result in part a) as a base case for an inductive argument. But I can't really figure out what to induct on, or how to do it. Any suggestions?
    I do not think what you wrote is true (the equation in (b) is wrong). This is because if t is a transposition it means that \text{sgn}(t) = (-1)^{n-1} also \text{sgn}(g) = (-1)^{n-k}. The second part of the exercise claims that \text{sgn}(tg) = \text{sgn}(t)\text{sgn}(g). But by the first part of exercise \text{sgn}(tg) = -\text{sgn}(g). This means that \text{sgn}(t)\text{sgn}(g) = - \text{sgn}(g). Canceling we get that \text{sgn}(t) = -1 however \text{sgn}(t) = (-1)^{n-1} and we we are saying that (-1)^{n-1} = (-1). This is not true unless n is an even number.

    (Even though this problem seems to be faulty the general idea outlined
    in my first post is one possible way to approach such a problem).
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  5. #5
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    Hmm.. well I know for a fact that the sgn function is a homomorphism, which is what part b) is asking to prove. Maybe my definition was wrong.. it would be fine if n was, instead of being the number of elements in the set, the number of elements in the cycle, wouldn't it? Because then that way, n for a transposition would be 2, and k would be 1, so sgn(t) = (-1)^(2-1) = -1. Alternatively, a definition of sgn that I know is correct is sgn(g) = 1 if g is an even permutation, and sgn(g) = -1 if g is an odd permutation. Now can you think of a way to do part b)?
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  6. #6
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    Ah ok, I clarified the definition of signum with my professor. n is still the size of the set but k is the number of cycles in the cycle decomposition of g, but including all of the 1-cycles too
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