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Math Help - Proof of intersect between 2 elements of group G

  1. #1
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    Proof of intersect between 2 elements of group G

    Hi All

    How do I prove the following:

    Suppose that G is a group with elements a and b of orders p and q respectively, where p and q are distinct primes.

    i) Show that <a> Intersect <b> = {e}, where e is identity element
    ii)Suppose that for some intergers i and j, a^i = b^j. Show that p|i and q|j

    Thanks again
    The moolmiester returns
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  2. #2
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    Is the first part of the question the same as the working in this one I found on the net?

    27. Let G, G1, G2 be groups. Prove that if G is isomorphic to G1 G2, then there are subgroups H and K in G such that
    (i) H K = { e },
    (ii) HK = G, and
    (iii) hk=kh for all h in H and k in K. Solution: Let : G1 G2 -> G be an isomorphism. Exercise 3.3.9 in the text shows that in G1 G2 the subgroups
    H* = { (x1,x2) | x2 = e } and K* = { (x1,x2) | x1 = e } have the properties we are looking for.
    Let H = (H*) and K = (K*) be the images in G of H* and K*, respectively. We know (by Exercise 3.4.15) that H and K are subgroups of G, so we only need to show that
    H K = { e }, HK = G, and hk=kh for all h in H and k in K.
    Let y be in G, with y = (x), for x in G1 G2. If y is in H K, then y is in H, and so x is in H*. Since y is in K as well, we must also have x in K*, so x is in H* K*, and therefore x = (e1,e2), where e1 and e2 are the respective identity elements in G1 and G2. Thus y = ((e1,e2)) = e, showing that H K = { e }. Since y is any element of G, and we can write x = h* k* for some h* in H* and some k* in K*, it follows that y = (h* k*) = (h*) (k*), and thus G = HK. It is clear that preserves the fact that elements of H* and K* commute. We conclude that H and K satisfy the desired conditions.
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  3. #3
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    Quote Originally Posted by moolimanj View Post
    i) Show that <a> Intersect <b> = {e}, where e is identity element
    ii)Suppose that for some intergers i and j, a^i = b^j. Show that p|i and q|j
    For the first question say x\in \left< a\right> \cap \left< b\right> then by Lagrange's theorem \mbox{ord}(x)|\mbox{ord}(a)|\mbox{ord}(a) and \mbox{ord}(x)|\mbox{ord}(b) but \mbox{ord}(a) = p and \mbox{ord}(b)=q so \mbox{ord}(x) divides both p and q thus \mbox{ord}(x) = 1\implies x=e because \gcd(p,q)=1.
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  4. #4
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    Thanks perfecthacker,

    Can you help with the second part as well?

    Cheers
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