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1) Is $\mathbb{Z}/\langle n\rangle$ cyclic and of order $n$? Why or why not? ($\mathbb{Z}/\langle n\rangle$ is defined to be the factor group of $\mathbb{Z}$ determined by $\langle n\rangle$.)

I guess I'm having trouble understanding factor groups (quotient groups make much more sense) and cyclic groups. My textbook isn't the best at explaining what's going on. I just chose some past homework problems that looked related to my problems.

This is my first abstract algebra course and I'm floundering a lot.

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    $\begingroup$ Try to prove this fact: The homomorphic image of a cyclic group is cyclic. $\endgroup$
    – Amr
    Sep 15, 2013 at 22:02
  • $\begingroup$ The group is generated by (among others) the equivalence class of $114. $\endgroup$ Sep 15, 2013 at 22:08
  • $\begingroup$ $5\mathbb{R}=\mathbb{R}$. The left hand side is clearly contained in the right hand side, and if $r\in \mathbb{R}$, then $r'=r/5$ is also a real, so $5r'=r\in 5\mathbb{R}$. $\endgroup$ Sep 15, 2013 at 22:27
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    $\begingroup$ What is the difference between "quotient groups" and "factor groups"? I don't mean this as a leading hint: I mean I actually have no idea what the difference is. Aren't those just different names for the same concept? $\endgroup$ Sep 16, 2013 at 1:13
  • $\begingroup$ I think so, from what I've gathered from other people online. My textbook uses "factor groups" when "quotient groups" makes a lot more sense. $\endgroup$ Sep 16, 2013 at 1:29

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From your notation and my limited knowledge of terminology, I would say that by a factor group one means the same as a quotient group.

So with your example of $\langle n\rangle = n\mathbb{Z} = \{\dots, -2n, -n , 0 , n, 2n , \dots\}$ we simply have that $$ \mathbb{Z}/ n\mathbb{Z} = \{[0], [1], \dots, [n-1]\}. $$ Here we indeed get an abelian groups under $[n]+ [m] = [n+m]$. Now you can probably figure out whether or not the group is cyclic.

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How does $[1]$ behave? I think that will tell all.

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