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Math Help - External Direct Product Question

  1. #1
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    External Direct Product Question

    Question:

    If p and q are odd primes and m and n are positive integers show why  U(p^m) \oplus U(q&n) is not cyclic

    where U(n) = { a < n | gcd(a,n)=1}

    Attempt:

    I know that the group is isomorphic to U(p^m*q^n) so if I can show that one is not cyclic I can get the answer.

    It must somehow involve the fact that p and q are odd

    so I said p=2k+1 and q = 2L+1 where k and and L are primes. Then I know the order of the group is (2k+1)^(m-1)*2k*(2k+1)^(n-1)*(2L)

    but I do not know where to go from there. Any tips?
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  2. #2
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    That's supposed to be q^n, not qn, correct?

    You can't express the orders of the primes like that... at least, not assuming k and L are prime. For example, k=6 gives you a prime.

    The orders of U(p^m),U(p^n) are \varphi(p^m)=p^{m-1}(p-1),\varphi(q^n)=q^{n-1}(q-1) respectively. So the direct product has order p^{m-1}q^{n-1}(p-1)(q-1). Also, notice that, since p,q are odd, each of p-1,q-1 is even.

    We can actually prove a stronger result here:

    If A,B are two groups such that |A|,|B| are not relatively prime, then the direct product A\times B cannot be cyclic.

    Notice that this would settle this problem as well. Can you prove this? (Consider the effect of raising an element of A\times B to the power of \mathrm{lcm}[|A|,|B|].)
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  3. #3
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    I can see how my problem is an example of the more general proposition you provided but I am still having trouble with the proof.

    Consider (a,b) an element of the direct product of A=U(p^m) and B=U(q^n)

    then (a,b)^lcm(|A|,|B|) = (a^(p^(m-1)*q^(n-1)*(p-1)*(q-1)) , b^(p^(m-1)*q^(n-1)*(p-1)*(q-1)) )

    also I see that |A| and |B| are NOT relatively prime (since p-1 and q-1 are even)
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  4. #4
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    I figured it out but I used a different (similar) method
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  5. #5
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    Oh okay, that's cool.

    Here's what I had in mind:

    Take an element from some group. If you raise this element to the power of the order of the group, you must get the identity, correct?

    It follows that if you raise the element to a power which is a MULTIPLE of the order of the group, you still get the identity.

    Now, in the direct product, the LCM of the two orders is a multiple of both individual orders. So if you raise any element to that power, you should get the identity in both components; that is, the identity element of the direct product.

    Thus any element of the direct product has order less than or equal to this LCM. Finally, it is a well known formula that LCM[m,n]=\frac{mn}{GCD(m,n)}. If m and n are not relatively prime, then clearly their LCM is strictly smaller than their product. But in our case, the LCM represents the maximal order of an element; the product is the order of the direct product. So no element may have order equal to the order of the group.
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