Math and science::Algebra::Aluffi
Groups. Order for commutative product.
Order for commutative product
If \( gh = hg \), then \( |gh| \) divides \( \operatorname{lcm}(\; |g|, \; |h| \;) \).
Proof
Let \( N = \operatorname{lcm}(\; |g|, \; |h| \; ) \).
\[
\begin{align}
(gh)^N &= g^Nh^N && \text{(by commutitivity)} \\
&= e \; e && \text{(as both $|g|$ and $|h|$ divide $N$)} \\
&= e
\end{align}
\]
By Lemma 1.10 in Aluffi, it follows that \( |gh| \) divides \( \operatorname{lcm}(\; |g|,\; |h| \;) \).
Example of not-equal
The proposition states "divides" and doesn't say "equal". Some examples where equality does not hold:
- If \( h = g^{-1} \), then \( |gh| = 1 \), regardless of \( |g| \).
- If \( g = [1]_{10} \) and \( h = [4]_{10} \) then \( ghgh = [0]_{10} \), so \( |gh| = 2 \), but \( |g| = 10 \) and \( |h| = 5 \).
Condition for equality
If \( g \) and \( h \) commute and \( \operatorname{gcd}(|g|, |h|) = 1 \), then \( |gh| = |g| |h| \).
Proof:
Is there a nice visualization for this?