$gh = Hg, \ \gcd(|g|, |h|) = 1\Rightarrow|gh - Math Stack Exchange

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Learn more about Teams $gh = hg, \ \gcd(|g|, |h|) = 1\Rightarrow|gh| = |g||h|\ \ (|a|$ = order of element $a)$ Ask Question Asked 8 years, 6 months ago Modified 4 years, 8 months ago Viewed 5k times 10 $\begingroup$

Let $G$ be a group and $g,h \in G$. I need to prove that if $g$ and $h$ commute and their orders are coprime, then $|gh| = |g||h|$, that is, the order of their product is the multiple of their orders.

Since $gh = hg$, then $(gh)^{ lcm(|g|, |h|)} = e$, so, $|gh|$ divides $lcm(|g|, |h|)$. Since $\gcd(|g|, |h|) = 1$, then $lcm(|g|, |h|) = |g||h|$.

So, $|gh|$ divides $|g||h|$.

• abstract-algebra
• group-theory

Share Cite Follow edited Sep 12, 2019 at 22:22 Bill Dubuque 278k4040 gold badges319319 silver badges976976 bronze badges asked Jun 22, 2016 at 21:16 Jxt921Jxt921 4,64822 gold badges2121 silver badges6565 bronze badges $\endgroup$ 0 Add a comment |

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You can find $|gh|$ divides $|g||h|$ already from $$(gh)^{|g||h|}=g^{|g||h|}h^{|h||g|}=e^{|h|}e^{|g|}=e.$$ On the other hand, $$ e=(gh)^{|gh||h|}=g^{|gh||h|}h^{|gh||h|}=g^{|gh||h|}$$ implies $|g|$ divides $|gh||h|$. As $\gcd(|g|,|h|)=1$, we conclude $|g|$ divides $|gh|$. Similarly, $|h|$ divides $|gh|$. Again using $\gcd(|g|,|h|)=1$, we finally have $|g||h|$ divides $|gh|$.

Share Cite Follow edited Apr 17, 2020 at 6:35 answered Jun 22, 2016 at 21:22 Hagen von EitzenHagen von Eitzen 379k3131 gold badges366366 silver badges666666 bronze badges $\endgroup$ 0 Add a comment | 4 $\begingroup$

$1 = (gh)^k\! = g^k h^k\Rightarrow\, g^k\! = h^{-k} =: \color{#c00}f\in \langle g\rangle\cap\langle h\rangle\Rightarrow\, |f|\ {\Large\vert}\ \overbrace{|g|,|h|}^{\rm coprime}\,\Rightarrow |f|=1\Rightarrow \color{#c00}{f = 1}$

So $\,\ (gh)^k = 1\iff g^k =\color{#c00} 1 = h^k\!\iff |g|,|h|\mid k\iff\! |g|\,|h|\mid k,\,$ thus $\,|gh| = |g||h|$

Share Cite Follow edited Jul 30, 2019 at 1:18 answered Jun 22, 2016 at 21:41 Bill DubuqueBill Dubuque 278k4040 gold badges319319 silver badges976976 bronze badges $\endgroup$ 3

• $\begingroup$ How do you get the last implication? If $|g|, |h| \mid k$ then $|g||h| \mid k$? $\endgroup$ – user5826 Commented May 14, 2019 at 2:47
• 1 $\begingroup$ @AlJebr By the LCM universal property, i.e. $\,a,b\mid k \iff {\rm lcm}(a,b)\mid k.\,$ Here $\,a,b = |g|,|h|\,$ are coprime $\gcd(a,b) = 1\,$ so $\,{\rm lcm}(a,b) = ab\,$ by here. Thus for coprime $\,a,b\,$ we have $\,a,b\mid k\iff ab\mid k\ \ \ $ $\endgroup$ – Bill Dubuque Commented May 14, 2019 at 3:00
• $\begingroup$ Compare here. $\endgroup$ – Bill Dubuque Commented Jul 23, 2019 at 21:52

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2 Order of a sum of elements in an abelian group 0 $|ab| =$ LCM $(|a|,|b|)$ in groups 22 How to represent "not an empty set"? 26 $a^n \equiv 1\pmod{\!m}\!\iff\! $ order of $a$ divides $n\,$ [Order Theorem] 5 Show that lcm$(a,b)= ab$ if and only if gcd$(a,b)=1$ 3 $ d_1,d_2\mid n\iff {\rm lcm}(d_1,d_2)\mid n\,\ $ [LCM Universal Property or Definition] 0 If $a$ and $b$ are elements of an abelian group with $(|a|,|b|)=1$, then $|ab|=|a||b|$.

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55 Examples and further results about the order of the product of two elements in a group 1 Existence of an Element with a particular order 10 Order of the product of two commuting elements with coprime orders in a group. 0 Sets of positive integers closed under lcm/gcd? 1 $\gcd \cdot \mathrm{lcm}$ for cyclic rings 0 find the order of the element $a^{-1}b?$ 6 Classifying finite groups where order is multiplicative on elements with coprime orders

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