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Let $G$ be an abelian group, $g_1,g_2\in G$ of finite order ($o(g_1)=m,o(g_2)=n)$ with $(o(g_1),o(g_2))=1$ (relatively prime). Show that $o(g_1\cdot g_2)=o(g_1)\cdot o(g_2)$.

I have tried the following:

$o(g_1)\cdot o(g_2)=m\cdot n$. Then, if $(g_1\cdot g_2)^{m\cdot n}=e\Rightarrow o(g_1\cdot g_2)=o(g_1)\cdot o(g_2)$.

$(g_1\cdot g_2)^{m\cdot n}=(g_1)^{m\cdot n}\cdot (g_2)^{m\cdot n}$ because $G$ is abelian. So, $(g_1)^{m\cdot n}\cdot (g_2)^{m\cdot n}=(g_{1}^{m})^n\cdot (g_{2}^{n})^m=e^n\cdot e^m=e\cdot e=e$

But I have not used the hypothesis that they are relatively prime.

Marco García
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2 Answers2

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Remember that the order is the smallest such power $n$ such that $g^n = e$. Hence, if the two numbers are not relatively prime, then the power becomes $lcm(m, n)$. Since they are prime, it translates to $m * n$.

Siddharth Bhat
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Write $l=o(a), k=o(b)$ and $m=o(ab)$. We have $$(ab)^{kl}=(a^{l})^{k}(b^{k})^{l}=e$$ thus $m|kl$. Now, note that $$a^{km}=(ab)^{km}=e \ \mbox{and}\ b^{lm}=(ab)^{lm}=e$$ thus $l|km$ and $k|lm$. Since $gcd(l,k)=1$ then $l|m$ and $k|m$. Therefore $kl=lcd(k,l)|m.$

Rafael
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