Let G be an abelian group. Suppose $a, b$ are elements of orders $m$ and $n.$ Let $d =\operatorname{lcm}(m, n).$ Show that $(ab)^d = 1$

Question

Here are my thoughts so far: G is abelian so $ab = ba.$

For $d = mn, (ab)^{mn} = abab..ab = aaaa..bbb$ as abelian $= a^{mn}b^{mn} = e$ as $a^m = e$ and $b^n = e.$

For $d< mn,$ let $d = qm$ and $d = pn.$ Then $$\begin{align}(ab)^{qm} &= a^{qm}b^{qm} \\& = eb^{qm} \\&= b^{qm}\\ (ab)^{pn} &= a^{pn}b^{pn} \\& = a^{pn}e \\&= a^{pn}\end{align}$$ so $b^{qm} = a^{pn}$ so $a^d = b^d.$

I'm not really sure how to continue from here/ if this is even the logically path I should be following

score 2 · Accepted Answer · answered Nov 02 '20 at 13:13

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So when $G$ is abelian, we have that $(ab)^d=a^db^d=a^{qm}b^{pn}=(a^m)^q(b^n)^p=e^qe^p=e\cdot e=e$.

answered Nov 02 '20 at 13:13

Let G be an abelian group. Suppose $a, b$ are elements of orders $m$ and $n.$ Let $d =\operatorname{lcm}(m, n).$ Show that $(ab)^d = 1$

1 Answers1