# |a\times b| = LCM[|a|, |b|]

• Sep 9th 2011, 08:20 PM
dwsmith
|a\times b| = LCM[|a|, |b|]
A and B are groups with binary operations $\displaystyle *$ and ?

Prove that the order of $\displaystyle a\times b$ is $\displaystyle \text{LCM}[|a|,|b|]$

The binary operation of $\displaystyle A\times B$ is define as such:
$\displaystyle (a\times b)(c\times d)=(a*c)\times (b\text{?} d)$

Let $\displaystyle |a|=n$ and $\displaystyle |b|=m$.

Do I need to define the gcd as well? Regardless if I do or not, I am not sure where to proceed next.
• Sep 9th 2011, 10:35 PM
fatpolomanjr
Re: |a\times b| = LCM[|a|, |b|]
LCM(n, m) is the smallest integer that is both a multiple of both n and m. By definition the order of $\displaystyle a\times b$ is the least such integer $\displaystyle k$ with $\displaystyle (a \times b)^k = (a*\cdots *a) \times (b*\cdots *b) = a^k \times b^k = e \times e$. By definition of the order of a and b, we know that $\displaystyle a^k = e \Leftrightarrow n | k$ and $\displaystyle b^k = e \Leftrightarrow m | k$. Therefore, k is the least such integer such that k = $\displaystyle l_an$ and k = $\displaystyle l_bm$ (these expressions come from definition of n "divides" k, or "n | k"). This means, overall, that k is the least such integer that is a multiple of both n and m.