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

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

A and B are groups with binary operations $*$ and ?

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

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

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

Do I need to define the gcd as well? Regardless if I do or not, I am not sure where to proceed next.

2. ## 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 $a\times b$ is the least such integer $k$ with $(a \times b)^k = (a*\cdots *a) \times (b*\cdots *b) = a^k \times b^k = e \times e$. By definition of the order of a and b, we know that $a^k = e \Leftrightarrow n | k$ and $b^k = e \Leftrightarrow m | k$. Therefore, k is the least such integer such that k = $l_an$ and k = $l_bm$ (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.