Is $\mathbb{Z}/\langle n\rangle$ cyclic and of order $n$?

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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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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.