Re: group problem
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 31 May 2005 16:50:08 +0000 (UTC)
In article <1117499897.682153.176780@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Li Yi <liyi.cn@xxxxxxxxx> wrote:
>Let G be an abelian group and suppose that G has elements of orders
>m and n, respectively. Prove that G has an element whose order is
>the least common mutiple of m and n.
Yes; this one comes up often in the newsgroup...
>The following proof is presented in 'problems from herstein'.
>=====================
>Proof. Let o(a) = m and o(b) = n. If (m, n) = 1, then let c = ab.
>Clearly, c^[m,n] = c^mn = a^mn b^mn = e. If o(c) = r, we have a^r =
>b^(-r). Thus, e = a^mr = b^(-mr). But then n|mr. Since (m, n) = 1,
>n|r. Similarly, m|r. This implies that [m, n] = mn|r and thus mn = r.
In other words: when m and n are relatively prime, the order of ab is mn.
>If (m, n) = d > 1, then o(a^d) = m/d and o(b^d) = n/d and (m/d, n/d) =
>1. Thus (ab)^d has order mn/d^2 by previous case and thus ab has order
>[m, n] = mn/d.
>=====================
>I wonder the last step. Why does ab have order mn/d if (ab)^d has the
>order of mn/d^2?
>
>If o(a^2)=5, o(a) is not necessary to be 10. It can be 5, too.
Ah, but the difference here is that 2 and 5 are relatively prime. mn/d
and d are not relatively prime. But you are right: the assertion does
not follow in general. If a=b^{-1}, then m=n=d, and the claim is that
ab has order d, which is false when d>1.
But the idea is on the right path. You want to take a power of a, a^q,
and a power of b, b^p, such that the orders of a^q and b^p are
relatively prime, and their product is equal to the gcd of m and
n. The easiest way to do that is to factor m and n into primes,
m = p_1^{a_1} * ... * p_r^{a_r} with a_i>=0 and all primes distinct
n = p_1^{b_1} * ... * p_r^{b_r} with b_i>=0 and all primes distinct
and then let q be the product of all p_j^{a_j} for which a_j<b_j, and
let p be the product of all p_j^{b_j} for which b_j<=a_j. Then each
p_i appears in only one of the orders for a^q and b^p, and it apears
raised to the max{a_i,b_i} power.
That's the classical proof. Or you can try to do it via the problem I
submitted to the Monthly some time ago whose answers have not yet been
printed (quoted from memory):
Let G be a group and let x and y be elements of G that commute with
one another. Let m be the order of x and let n be the order of
y. It is well known that if m and n are relatively prime, then the
order of xy is mn. Prove the following generalization: if for every
prime p that divides mn, the largest power of p that divides n is
different from the largest power of p that divides m, then the
order of xy is lcm(m,n).
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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