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I am interested in getting a stronger intuition for quotient groups. I found this related thread on MSE, but these answers do not seem sufficient for me.

The way I look at quotient groups is the following:

If we have a group $G/H$, then we split up the group $G$ in bigger parts, with the same size as $H$. Moreover, $H$ becomes the neutral element of this group.

For example, when we consider the quotient $(\mathbb{R}_0,.)/(\mathbb{R_0}^{+},.)$, it makes sense that this group is isomorphic to $\mathbb{Z}_2$, since we divide $\mathbb{R_0}$ in 2 bigger parts ($\mathbb{R_°^{+}},-\mathbb{R_°^{+}})$

Another example: $\mathbb{Z}/3\mathbb{Z}$

Here, in the quotient,we have that $3\mathbb{Z}$ will become the neutral element of the quotient group. This means, every time we encounter a multiple of $3$ in $\mathbb{Z}$, we start over again and counting from $0$.

However, I find that my intuition is lacking when considering examples like $\mathbb{R}/\mathbb{Z}$, which should be isomorphic to the unit circle in the complex plane.

So, are there other ways to visualise/think about quotient groups?

Thanks in advance.

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    A stronger intuitioon, I think, is only possible with additional arguments - so see here for your last example. – Dietrich Burde Jun 22 '17 at 09:30
  • Normal subgroups always appear as the kernel of some group morphism. Understanding the group morphism helps you to understand the quotient. Indeed, for the last example the map $\phi:\mathbb{R}\rightarrow \mathbb{C}:x\mapsto e^{2\pi i x}$ has $\mathbb{Z}$ as it's kernel. Thus the first isomorphism theorem says that $\mathbb{R}/\mathbb{Z}\cong \text{im}(\phi)$. The point is that the first isomorphism theorem is your friend when it comes to quotients. Finding the proper map might be more natural than thinking about the quotient directly. – Mathematician 42 Jun 22 '17 at 09:35
  • @Mathematician42 Well noticed. This question is more about a general intuition though. The last example was just an illustration where intuition could lack. –  Jun 22 '17 at 09:38
  • http://www.math3ma.com/mathema/2016/10/17/whats-a-quotient-group-really-part-1 this is one of the most intuitive explanations of quotient groups I've seen, check out part 2 as well – Osama Ghani Jun 22 '17 at 09:40
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    I thought $\Bbb R/\Bbb Z\cong S^1$ is obvious, if you include the topology. Anyway, quotient groups are not always easy to visualise. $\Bbb R/\Bbb Q$ is an example. – edm Jun 22 '17 at 09:43
  • Consider $\mathbb{R}[X]/(X^2-1)$. I always think of what you kill in a quotient and what survives. So here we kill $X^2-1$, thus $X^2=1$. Clearly $\mathbb{R}[X]/(X^2-1)$ is "larger" than $\mathbb{R}$ since $X^1$ survives. So actually you are adjoining a solution of $X^2-1=0$ to $\mathbb{R}$. Unsurprisingly, the first isomorphism theorem and the map $\phi:\mathbb{R}[X]\rightarrow \mathbb{C}:p(X)\mapsto p(i)$ give that $\mathbb{R}[X]/(X^2-1)\cong \mathbb{C}$. Off course these are quotient of rings, but the idea is the same. Polynomial rings and quotient are very important to algebraic geomtery. – Mathematician 42 Jun 22 '17 at 09:55
  • $\mathbb{R}[X]/(X^2+1)\cong \mathbb{C}$ – Lozenges Jun 22 '17 at 10:24
  • I see it as follows: if $N$ is normal in $G$ then $G/N$ is $G$ with the additional rule that "$N={e}$", so that $N$ is forced to be trivial. – M. Van Jun 22 '17 at 11:14
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    @Mathematician42 : you meant $X^2 +1$. Math_QED: as many have already said on this site, and as M.Van explains, quotienting is a bit like expanding the definition of equality : in $G/N$, you expand it so that for $n\in N$, $n=e$ is in the expansion. – Maxime Ramzi Jun 22 '17 at 12:24
  • I'm not sure, but you may also want to look at this thread. – Jyrki Lahtonen Jun 22 '17 at 17:38

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A natural way to visualize the isomorphism $\mathbb{R}/\mathbb{Z} \cong S^1 $, induced from the epimorphism $ p \colon \mathbb{R} \to S^1 $ defined by $ p(t) = e^{2\pi it}$ for $ t \in \mathbb{R} $ with $ \ker p = \mathbb{Z} $, is the following (though which does not work well in general).

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