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I have one simple question regarding quotient groups.

If there is a quotient group which consists of all remainders when an integer is divided by 12 such group could be considered as a cyclic group generated by a set of integers congruent to 1 or 5 or 11 mod 12.

My problem: As far as I can understand ( and I would not be surprised if I was totally wrong) cyclic group consisting of all numbers congruent to 1 mod 12 will not include all integers which give remainder 0, hence how such a cyclic group can be considered equal to a quotient group which consists of all remainders when divided by 12.

Misc
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  • I'm having a little trouble parsing the second sentence. It might just be me but it seems like it is missing a word. It couldn't hurt to split it up into several smaller sentences if possible. – Sriotchilism O'Zaic Oct 17 '17 at 18:35

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The group is $\mathbb{Z}/12\mathbb{Z}$. So its elements are cosets $0+12\mathbb{Z},\ldots, 11+12\mathbb{Z}$. It is generated by, say $a=1+12\mathbb{Z}$. Note however, that generator means something different than than what you say. We have $\langle a\rangle\neq \{a\}$ here.

Dietrich Burde
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  • I am confused because I think that cyclic group Z/12Z is generated by 1+12Z same way Z is generated by 1. I hope this makes sense. – Misc Oct 17 '17 at 18:57
  • You can also take $a=5+12Z$. Then $2a=10+12Z$, $3a=15+12Z=3+12Z$, etc. So it is not the same way as in $Z$, because $5$ does not generate $Z$. – Dietrich Burde Oct 17 '17 at 19:01
  • Hmm, it appears that I simply have no idea what "generated" means in this context... Would it be too much to ask some info on that, sorry if I am wasting your time. – Misc Oct 17 '17 at 19:02
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    No problem. But like reading a text in a foreign language, just open the dictionary yourself. A good place to start is here. – Dietrich Burde Oct 17 '17 at 19:05
  • Thanks for your time. – Misc Oct 17 '17 at 19:09