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  1. #1
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    Let G be a group and p \in N^* s.t equation x^p=e have unique solution . Prove that for every  a\in G the equation x^p=a have unique solution
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    Quote Originally Posted by petter View Post
    Let G be a group and p \in N^* s.t equation x^p=e have unique solution . Prove that for every  a\in G the equation x^p=a have unique solution
    Suppose x^p= a and y^p= a. What can you say about (xy^(-1))^p?
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    Quote Originally Posted by HallsofIvy View Post
    Suppose x^p= a and y^p= a. What can you say about (xy^(-1))^p?
    But does this only work if G is assumed to be abelian?

    ---
    Is p a prime? I will assume that p is a prime.

    We will prove this for S_n i.e. if x^p = y^p \implies x=y where x,y\in S_n. Now say we can write x = zx_1 and y=zy_1 where z is a disjoint permutation with x_1 and y_1. This means z commutes with x_1 and y_1. Thus, x^p = z^px_1^p and y^p = z^py_1^p henceforth x_1^p = y_1^p. We can write x = \sigma_1 ... \sigma_r as a product of disjoint cycles and y=\tau_1 ... \tau_s as a product of disjoint cycles. We want to show x=y. Based on the reasoning above we can assume that each \sigma_i and \tau_j are different. It follows that x^p = \sigma_1^p ... \sigma_r^p and y^p = \tau_1^p ... \tau_s^p. Since \sigma_i^p,\tau_j^p are non-identity elements it means we have expressed x^p as a product of disjoint cycles (for \sigma_i^p,\tau_j^p are cycles - this is because p is a prime!). Therefore, not only r=s but it must be that we can rearrange them so that \sigma_i^p = \tau_i^p. Note, it is impossible that \sigma_i \not = \tau_i and \sigma_i^p = \tau_i^p. This means that x=y. Now by the Cayley's theorem and finite group and be embedded in the symmetric group. Since this result applies to symmetric groups it will apply to its subgroups as well. Thus, this is true for any finite group G and a prime p.

    What if p is not a prime?
    Perhaps, there is a conterexample - but I am too lazy to try to find one now.
    Last edited by ThePerfectHacker; October 25th 2008 at 03:56 PM.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    What if p is not a prime?
    Perhaps, there is a conterexample - but I am too lazy to try to find one now.
    If p is not a prime then this statement is still true.

    Say x^n = e has exactly one solution (trivial one) and n>1. Then it must be that x^p = e has exactly one solution for each p|n a prime. Now if x^n = y^n it means (x^{(n/p)})^p = (y^{(n/p)})^p \implies x^{(n/p)} = y^{(n/p)} (by the above post). If n/p is prime then it would be that x=y. Otherwise let q be prime divisor of (n/p) and arrive at x^{n/(pq)} = y^{n/(pq)}. Keep on doing this until you exhaust all the primes and arrive that x=y.
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