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Math Help - Group of Units (Abelian)

  1. #1
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    Group of Units (Abelian)

    I am in dire need of help! U_n is an abelian group under multiplication mod (n). If you need more information, please let me know.

    1) Find an element of maximal order in U_{884}.

    2) Let a and e be integers with e \geq4 and |[a]_{16}|=4. Prove that |[a]_{2^e}|= 2^{e-2}

    3) Let e and f be positive integers.
    (a) Determine the number of elements of order 2^f in U_{2^e}.

    (b) Compute |\{a^{2^f} | a\in U_{2^e}\}|

    Thanks for any and all help! I need help with these by tomorrow night, thanks
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    I assumed that U_n is the group of all invertible congruence classes modulo n under multiplication.

    Quote Originally Posted by shadow_2145 View Post
    1) Find an element of maximal order in U_{884}.
    Note that 884 = 4\cdot 13\cdot 17.
    Thus, U(\mathbb{Z}_{884})\simeq U(\mathbb{Z}_4)\times U(\mathbb{Z}_{13})\times U(\mathbb{Z}_{17})\simeq \mathbb{Z}_2\times \mathbb{Z}_{12}\times \mathbb{Z}_{16}.
    The element ([1]_2,[1]_{12},[1]_{16}) has maximum order with \text{lcm}(2,12,16)=48.
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    Quote Originally Posted by shadow_2145 View Post
    2) Let a and e be integers with e \geq4 and |[a]_{16}|=4. Prove that |[a]_{2^e}|= 2^{e-2}
    |[a]_{16}| = 4 means a^4 \equiv 1(\bmod 16), another way to write this is as a^{2^2} \equiv 1(\bmod 2^4).
    We need to prove that a^{2^{e-2}}\equiv 1(\bmod 2^e) we do this by induction.
    If e=4 there is nothing to prove because that is already given to us.
    Assume that a^{2^{k-2}} \equiv 1(\bmod 2^k) holds for k\geq 4.
    Then, \left( a^{2^{k-2}} \right)^2 \equiv 1(\bmod 2^{k+1}) thus, a^{2^{k-1}} \equiv 1(\bmod 2^{k+1}) completing induction.

    Note: We use the fact that if a\equiv b(\bmod p^n) then a^p \equiv b^p (\bmod p^{n+1}) where p is a prime.
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    Quote Originally Posted by shadow_2145 View Post
    3) Let e and f be positive integers.
    (a) Determine the number of elements of order 2^f in U_{2^e}.
    Remember a simple fact about cyclic groups. Let m\geq 2 and [k]_m \in \mathbb{Z}_m then the order of [k]_m = m/\gcd(k,m).

    Say that e\geq 3 (when e=1,2 you can do yourself). Then it is known that U(\mathbb{Z}_{2^e}) \simeq \mathbb{Z}_{2^{e-2}}\times \mathbb{Z}_{2}.
    Any element in \mathbb{Z}_{2^{e-2}}\times \mathbb{Z}_2 has form either: ([a]_{2^{e-2}},[0]_2) or ([a]_{2^{e-2}},[1]_2).
    The order of ([a]_{2^{e-2}},[0]_2) is simply the order of [a]_{2^{e-2}} in \mathbb{Z}_{2^{e-2}}.
    The order of ([a]_{2^{e-2}},[1]_2) is simply the order of [a]_{2^{e-2}} if [a]\not = [0], otherwise the order is 2.

    You are trying to find how many elements have order 2^f. Well, if f=0 then only ([0]_{2^{e-2}},[0]_2) has order 2^0 =1.
    If f=1 then ([0]_{2^{e-3}},[1]_2) has order 2^1=2 and all 1\leq a < 2^{e-2} so that \gcd(a,2^{e-2}) = 2^{e-3} have order 2.
    The only such a which makes this true is a=2^{e-3}. Thus for f=1 there are two elements with order 2^1.

    Now we will consider what happens when f\geq 2.
    The elements ([a]_{2^{e-3}},[0]_2) and ([a]_{2^{e-3}},[1]_2) so that \gcd( a,2^{e-2}) = 2^{e-f-2} with 1\leq a < 2^{e-2} are precisely that have this order.

    It remains to count the # of 1\leq a < 2^{e-2} so that \gcd(a,2^{e-2})=2^{e-f-2}.
    Note 2^{e-f-2}|a thus a=2^{e-f-2}b where 1\leq b< 2^f because we need 1\leq a < 2^{e-f-2}.
    Thus, \gcd(2^{e-f-2}b,2^{e-2}) = 2^{e-f-2}\implies \gcd(b,2^f) = 1 where 1\leq b < 2^f.
    There are of course \phi (2^f) = 2^{f-1} such numbers.

    Finally we need to double this number because that is the number for each case: ([a]_{2^{e-2}},[0]_2),([a]_{2^{e-2}},[1]_2).
    We get that the number of elements of order 2^f is equal to 2\cdot 2^{f-1} = 2^f.

    As a check note that when f=0 we should have 2^0 =1 element of order 1.
    Also when f=1 we should have 2^1=2 elements of order 2 (as we got above).
    Thus, eventhough this formula was derived for f\geq 2 it works for f\geq 0 as well.

    We conclude that there are 2^f elements of order 2^f.
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