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

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

$\displaystyle x^ny^m=e$

What is the commutativity used for?

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

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

$\displaystyle x^ny^m=e$

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

by the way, what you are trying to prove is false, without further conditions. consider m = n, and $\displaystyle y = x^{-1}$. - Oct 8th 2011, 07:11 PMdwsmithRe: LCM[|x|,|y|]
- Oct 8th 2011, 07:18 PMDevenoRe: 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 AMdwsmithRe: LCM[|x|,|y|]
Ok so here is what I have.

$\displaystyle x^ky^k=e$

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

$\displaystyle q_1,q_2\in\mathbb{Z}, \ 0\leq r_1 <n, \ 0\leq r_2<m$

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

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

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

Can I just say k is a common multiple and by definition of LCM, k|LCM? - Oct 9th 2011, 12:19 PMzoekRe: LCM[|x|,|y|]
- Oct 9th 2011, 12:27 PMzoekRe: LCM[|x|,|y|]
Let $\displaystyle k = LCM[m,n]$.

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

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