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I have an abstract algebra exam tomorrow, and I'm having a little bit of difficulty deciphering the difference between $\mathbb Z/6 \mathbb Z$, $\mathbb Z_6$ and $6\mathbb Z$.

Can someone please explain this to me? As far as I thought, $\mathbb Z_6$ was the cyclic group of order $6$ and $\mathbb Z/6\mathbb Z$ was the quotient group.... but in an assignment, my professor wrote that $\mathbb Z/6\mathbb Z = \mathbb Z_6$, so I just wanted to clarify

Also, please excuse my formatting here. I'm new to this website!

user5826
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ConfusedSoul
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  • For some basic information about writing math at this site see e.g. here, here, here and here. –  Oct 27 '14 at 00:57
  • I am going to assume that by "Z6", you mean $\mathbb{Z}_6$, which is indeed the cyclic group of order 6. In addition, $\mathbb{Z}/6\mathbb{Z}$ is also the cyclic group of order 6, where you take integers "modulo 6" (it is generated by the image of 1 in the quotient). As for $6\mathbb{Z}$, this is the additive subgroup of $\mathbb{Z}$, obtained by multiplying everything by $6$. – John Martin Oct 27 '14 at 00:59
  • I see that you (sort of) fixed the "Z6" thing. – John Martin Oct 27 '14 at 01:00
  • I'm sorry about the editing! I went to the links Chantry included, but I couldn't figure out how to fix it. – ConfusedSoul Oct 27 '14 at 01:04
  • Thanks for your help John! – ConfusedSoul Oct 27 '14 at 01:05

1 Answers1

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$\mathbb{Z}_6$ is the integers modulo $6$, as you know. $\mathbb{Z}/6\mathbb{Z}$ is the integers modulo the (normal) subgroup generated by $6$. They are the same group. To see this, just define define the homomorphism from the second to the first by $x+<6>\leadsto x \rm{mod} 6$, and look at the kernel then use the first isomorphism theorem. To check that this is well defined you will need to use the division algorithm.

This should not be suprising as two integers are in the same coset if and only if they have the same remainder modulo $6$. Also note that this is pretty much a tautology.

RougeSegwayUser
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  • So, just to make sure I'm getting this, ℤ6 = {0,1,2,3,4,5}, correct? And because ℤ/6ℤ is equal to {6ℤ, 6ℤ+1... 6ℤ+5}, they're the same thing? – ConfusedSoul Oct 27 '14 at 01:03
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    Well, I'm guessing that you're treating $\mathbb{Z}_6$ as a collection of formal symbols with the operation of addition modulo $6$. The difference isn't really important. The two aren't \emph{equal}, they are \emph{isomorphic}. – RougeSegwayUser Oct 27 '14 at 01:24