# LCM[|x|,|y|]

• Oct 8th 2011, 06:56 PM
dwsmith
LCM[|x|,|y|]
Assume |x| = n and |y| = m. Suppose that xy = yx. Prove that $|xy||\text{LCM}[n,m]$.

$\text{LCM}[n,m]=\frac{nm}{\text{gcd}(n,m)}$

$x^ny^m=e$

What is the commutativity used for?
• Oct 8th 2011, 07:00 PM
Deveno
Re: LCM[|x|,|y|]
to establish that $(xy)^k = x^ky^k$

by the way, what you are trying to prove is false, without further conditions. consider m = n, and $y = x^{-1}$.
• Oct 8th 2011, 07:11 PM
dwsmith
Re: LCM[|x|,|y|]
Quote:

Originally Posted by Deveno
to establish that $(xy)^k = x^ky^k$

by the way, what you are trying to prove is false, without further conditions. consider m = n, and $y = x^{-1}$.

That would work. The order of |xy| = 1. LCM[m,n] = n.

Then 1|n true.
• Oct 8th 2011, 07:18 PM
Deveno
Re: LCM[|x|,|y|]
my bad...didn't see the extra vertical line. in any case, what you do is show that (xy)^lcm(m,n) = e, and you're done!
• Oct 9th 2011, 08:46 AM
dwsmith
Re: LCM[|x|,|y|]
Ok so here is what I have.

$x^ky^k=e$

$k=nq_1+r_1 \ \text{and} \ k=mq_2+r_2$

$q_1,q_2\in\mathbb{Z}, \ 0\leq r_1

$x^k=\cdots x^{r_1} \ \text{and} \ y^k=\cdots y^{r_2}$

But m and n are least so $r_1,r_2=0$.

So we have $n|k \ \text{and} \ m|k$

Can I just say k is a common multiple and by definition of LCM, k|LCM?
• Oct 9th 2011, 12:19 PM
zoek
Re: LCM[|x|,|y|]
Quote:

Originally Posted by dwsmith
Ok so here is what I have.

$x^ky^k=e$

$k=nq_1+r_1 \ \text{and} \ k=mq_2+r_2$

$q_1,q_2\in\mathbb{Z}, \ 0\leq r_1

$x^k=\cdots x^{r_1} \ \text{and} \ y^k=\cdots y^{r_2}$

But m and n are least so $r_1,r_2=0$.

So we have $n|k \ \text{and} \ m|k$

Can I just say k is a common multiple and by definition of LCM, k|LCM?

By definition of LCM we have that LCM divides every commom multiple, so you can say that LCM/k - not k/LCM.
• Oct 9th 2011, 12:27 PM
zoek
Re: LCM[|x|,|y|]
Quote:

Originally Posted by dwsmith
Assume |x| = n and |y| = m. Suppose that xy = yx. Prove that $|xy||\text{LCM}[n,m]$.

Let $k = LCM[m,n]$.

Then $m/k$ and $n/k$. So, there exist $\lambda_1, \lambda _2$ such that: $k = \lambda_1 m$ and $k = \lambda_2 n$.

Now, $(xy)^k \overset{xy=yx}{==} x^k y^k = (x^n)^{\lambda_2} (y^m)^{\lambda_1}=e^{\lambda_2} e^{\lambda_1} = e \implies |xy|/k$.