2
$\begingroup$

Let $G$ be a finite abelian group and $d=o(ab), \ m=o(a), \ n=o(b)$.

Show that $d\mid \frac{mn}{\text{gcd}(m,n)}$ and $\frac{mn}{\text{gcd}(m,n)^2}\mid d$.

In particular, if $m$ and $n$ are coprime then order of product is multiplicative.

Proof: $(ab)^{\text{lcm}(m,n)}=a^{\text{lcm}(m,n)}b^{\text{lcm}(m,n)}=e$ then $d\mid \text{lcm}(m,n)$ or $d\mid \frac{mn}{\text{gcd}(m,n)}$. We have done with the first relation.

Since $e=(ab)^d=(ab)^{\text{gcd}(m,n)d}=a^{\text{gcd}(m,n)d}b^{\text{gcd}(m,n)d}$. If I'll show that $a^{\text{gcd}(m,n)d}=e$ and $b^{\text{gcd}(m,n)d}=e$ then $m\mid \text{gcd}(m,n)d$ and $n\mid \text{gcd}(m,n)d$ so we get what we need, i.e. $\frac{mn}{\text{gcd}(m,n)^2}\mid d$.

But as you see I have difficulties with showing that $a^{\text{gcd}(m,n)d}=e$.

Can anyone help with that, please?

$\endgroup$
2
$\begingroup$

If $ab$ has order $d$ in abelian group, then $(ab)^d=a^db^d=e$ so $a^d=b^{-d}$. Now write $(m, n)=ms+nt$ for some integers $s$ and $t$. So $$a^{(m, n)d}=a^{msd}\cdot a^{ntd}=b^{-ntd}=e.$$ Therefore $m|(m, n)d$. Similarly $n|(m, n)d$.

$\endgroup$
  • $\begingroup$ Thanks for help! Nice proof :) $\endgroup$ – ZFR Jan 17 '18 at 13:18
  • $\begingroup$ You are welcome! $\endgroup$ – PJK Jan 17 '18 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.