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I'm studying abstract algebra alone as my doctor needs a doctor haha. Ok, I'm studying about cyclic groups now. I read that:

if 2 elements $x$ and $y$ commute in a group $G$, then the order of $xy$ is a divisor of $l.c.m(o(x),o(y))$,

then I read that it's exactly equal to $l.c.m(o(x),o(y))$, so got confused. Can anybody explain this to me? Also, can anybody advise me to read some detailed lectures about cyclic groups, as I feel like if I need much more to be strong at this lesson.

Nour
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See the answer which was accepted from the other post.

Chickenmancer
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  • So you are telling me that $|xy|$ is exactly $lcm(x,y)$? if $G$ has infinite order then the identity is the only element that has finite order which's $1$. Isn't right? – Nour Apr 29 '17 at 16:15
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    In the case where $xy=yx,$ yes. If $G$ has infinite order, there can still be things of finite order. Take, for example, $GL_2(\Bbb{R})$ the group of $2\times 2$ matrices with real entries. This group is infinite in order, and has many elements of infinite order, but there are elements of finite order. An example of an element of order $2.$ $$\left(\begin{array}{cc} 0&1 \ 1&0 \end{array} \right)$$ – Chickenmancer Apr 29 '17 at 16:20
  • Mmmm, I can conclude that in cyclic groups only, if $G$ has infinite order then its $non identity$ elements do. Ok so if the group in general is not abelian then $|xy|$ is a divisor of $lcm(|x|,|y|)$? – Nour Apr 29 '17 at 16:24
  • Yes the abelian condition is crucial. Another example is in the dihedral groups. Take the principle rotation, and multiply it by any reflection, it now has order 2. – Chickenmancer Apr 29 '17 at 16:32
  • Ok, you really made a difference. Thank you :) – Nour Apr 29 '17 at 16:33
  • No trouble at all! Would you mind clicking the check mark under my answer? An up vote wouldn't hurt either! :) – Chickenmancer May 07 '17 at 04:26
  • @Chickenmancer Part of what you say here is not correct. If $xy=yx$, $o(x)=n$, and $o(y)=m$, then it is not necessarily true that $o(xy)=lcm(n,m)$. For example, if $G=\mathbb{Z}/6\mathbb{Z}$, $x=1$, and $y=3$, then $o(x)=6$, $o(y)=2$, and $o(xy)=3$. – Mr. Frog Aug 01 '17 at 20:16
  • @Chickenmancer This is still incorrect. For example, if $G=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$, $x=(1,0)$ and $y=(0,1)$, then $o(x)=2$, $o(y)=3$, and $o(xy)=6$ (but $gcd(2,3)=1$). It is clear that $o(xy)$ always divides $lcm(o(x),o(y))$, but in general there are multiple possibilities for $o(xy)$. – Mr. Frog Aug 06 '17 at 16:08