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Math Help - order of an element in mulitpicative group

  1. #1
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    order of an element in mulitpicative group

    This problem is from Abstract Algebra 3rd edition Dummit and Foote.

    Let p be an odd prime and n a positive integer. Use the Binomial Theorem to show that:

    (1+p)^{p^{n-1}} = 1 (\mbox{mod } p^n)

    but

    (1+p)^{p^{n-2}} \ne 1 (\mbox{mod } p^n)

    Deduce that that 1+p is an element of order p^{n-1} in the multiplicative group (\mathbb{Z}/p^n\mathbb{Z} \mbox{ }\cdot)

    Thanks

    As hint goes by the binomail theorem I get

    (1+p)^{p^{n-1}}=1+\sum_{i=1}^{p^{n-1}}\binom{p^{n-1}}{i}p^i=1+\sum_{i=1}^{p^{n-1}}\left(\frac{p^{n-1}!}{(p^{n-1}-i)!(i)!} \right)p^i

    What I have been trying from here (with no sucess) is to show that p^{n-1}|\left(\frac{p^{n-1}!}{(p^{n-1}-i)!(i)!} \right)p^i
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  2. #2
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    Quote Originally Posted by TheEmptySet View Post
    This problem is from Abstract Algebra 3rd edition Dummit and Foote.

    Let p be an odd prime and n a positive integer. Use the Binomial Theorem to show that:

    (1+p)^{p^{n-1}} = 1 (\mbox{mod } p^n)

    but

    (1+p)^{p^{n-2}} \ne 1 (\mbox{mod } p^n)

    Deduce that that 1+p is an element of order p^{n-1} in the multiplicative group (\mathbb{Z}/p^n\mathbb{Z} \mbox{ }\cdot)

    Thanks

    As hint goes by the binomail theorem I get

    (1+p)^{p^{n-1}}=1+\sum_{i=1}^{p^{n-1}}\binom{p^{n-1}}{i}p^i=1+\sum_{i=1}^{p^{n-1}}\left(\frac{p^{n-1}!}{(p^{n-1}-i)!(i)!} \right)p^i

    What I have been trying from here (with no sucess) is to show that p^{n-1}|\left(\frac{p^{n-1}!}{(p^{n-1}-i)!(i)!} \right)p^i
    The result is actually stronger, it goes that: (1+ap)^{p^{n-2}} \equiv 1 + ap^{n-1} (\bmod p^n).
    Where, n\geq 2, p>2, a\in \mathbb{Z}.

    We will prove this by induction.
    When n=2 there is nothing to prove.
    So say that, (1+ap)^{p^{n-2}} \equiv 1 + ap^{n-1}(\bmod p^n).
    We will prove it for n+1.

    Here is a useful result: if a\equiv b (\bmod p^k) \implies a^p \equiv b^p (\bmod p^{k+1})*.

    With this result we get, (1+ap)^{p^{n-1}}\equiv (1+ap^{n-1} )^p (\bmod p^{n+1} ).

    However, (1+ap^{n-1})^p \equiv 1 + ap\cdot p^{n-1} + \sum_{j=2}^p {p\choose j} a^j p^{j(n-1)} ... [1]

    Here is another result: if 1\leq j\leq p-1 then p divides {p\choose j}.**

    Thus every term in sum of [1] for j=2,...,p-1 is divisble by p^{1 + 2(n-1)} notice that 1+2(n-1) \geq n+1.
    Therefore, each term for j=2,...,p-1 is divisible by p^{n+1}. For the last term ( j=p) in sum of [1] we have a^p p^{p(n-1)}. But p(n-1) \geq n+1 since p>3. And so we have shown that p^{n+1} divides each term for j=2,...,p in the sum of [1]. Thus, the sum is equivalent to 0 mod p^{n+1}.

    Thus, (1+ap^{n-1})^p \equiv 1 + a p^n (\bmod p^{n+1}).

    *)Argue by induction. If a\equiv b(\bmod p^k) it means a = b + qp^k for some q\in \mathbb{Z}. Now raise both sides to the p power and apply the binomial theorem again. I think the result that appears in ** (just below) will be useful in this proof.

    **)Because {p\choose j} = \frac{p(p-1)...(p-j+1)}{j!} is an integer, since \gcd(j!,p)=1 for 1<j<p we have that j! divides (p-1)...(p-j+1). Thus, it leaves a factor of p. Thus, p divides {p\choose j}.

    ------
    We can now prove a stronger result. That the order of 1+ap is p^{n-1} mod p^n. Where p>2,p\not | a, n\geq 2.

    Okay we have shown (1+ap)^{p^{n-1}} \equiv 1 + ap^n (p^{n+1}). This implies, (1+ap)^{p^{n-1}} \equiv 1 (\bmod p^n). But, (1+ap)^{p^{n-2}} \equiv 1 + ap^{n-2} (\bmod p^n) (use binomial theorem). However, 1+ap^{n-2}\not \equiv 1 (\bmod p^n) since p\not | a. The rest follows.
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    Thanks for that awsome proof. It was very informative and the generalzation was awsome.
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