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Suppose that $G$ is a group and $N$ is a normal subgroup of $G$. How do we find out what $G/N$ is isomorphic to? For example, $C_m$ is a normal subgroup of $D_{2n}$ generated by a rotation of angle $2\pi/m$. We can partiotion the group $D_{2n}$ using the normal subgrouop $C_m$. We want to find a surjective homomorphism $\phi: G\to G'$ such that $\ker(\phi) = C_m$? What surjective homomorphism $\phi: G\to G'$ should we have so that $D_{2n}/C_m$ is isomorphic to $G'$?

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    $\begingroup$ Its difficult to work out what your question is. So far as I can tell, any answer is just going to re-state the first isomorphism theorem (and the current two answers do this). $\endgroup$
    – user1729
    Apr 2 '19 at 16:37
  • $\begingroup$ (On the other hand, the answer to the abstract question "how do we find out what $G/N$ is isomorphic to?" does have a concrete answer. The answer is: "you don't; in fact you can't even determine if $G/N$ is the trivial group".) $\endgroup$
    – user1729
    Apr 2 '19 at 16:41
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I assume $m|n$, so $n=md$ and $C_m\cong\langle r^d\rangle$. In this case, $|D_{2n}/C_m|=2d$. Working with cosets you can check that the quotient is generated by $[r]=rC_m$ and $[j]=jC_m$ satisfying the relations $[r]^d=[1]=C_m$, $[j]^2=[1]=C_m$ and $[j][r]=[r^{-1}][j]$. The correct guess is that $$D_{2n}/C_m\cong D_{2d}=\langle R,J\mid R^d=J^2=1,JR=R^{-1}J\rangle.$$

Now, define a homomorphism $\phi:D_{2n}\to D_{2d}$ by $\phi(r)=R$ and $\phi(j)=J$ (and extend multiplicatively). Check that this map is well defined (this involves the fact that $d|n$) and surjective. By the first isomorphism theorem $$D_{2n}/\ker\phi\cong D_{2d}.$$ Finally, check that $\ker\phi=\langle r^d\rangle$.

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  • $\begingroup$ Is it possible that $D_{2n}/C_m \cong$ some cyclic group? $\endgroup$
    – user546106
    Apr 2 '19 at 16:49
  • $\begingroup$ When $m=n$, it is isomorphic to $D_2=C_2$. Otherwise, the quotient is non-abelian. $\endgroup$
    – David Hill
    Apr 2 '19 at 16:50
  • $\begingroup$ Why is it a natural choice to pick $[r]=rC_m$ as a generator? What about $r^2C_m$? $\endgroup$
    – user546106
    Apr 2 '19 at 17:04
  • $\begingroup$ Because $r$ is a generator of $D_{2n}$ while $r^2$ is not, in general. $\endgroup$
    – David Hill
    Apr 2 '19 at 17:08
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Define the natural homomorphism $\gamma: G \to G/N$ by $g \mapsto gN.$ Then $N$ is the kernel of $\gamma$ and we have $$G/N\cong \operatorname{Im}(\gamma).$$

See Theorem 10.4 of Gallian's "Contemporary Abstract Algebra."

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  • $\begingroup$ Are you not saying something like $G/N\cong G/N$? I think $\operatorname{Im}(\gamma) = G/N$ $\endgroup$
    – user546106
    Apr 2 '19 at 16:28
  • $\begingroup$ No, since I'm applying the first isomorphism theorem, @user546106. $\endgroup$
    – Shaun
    Apr 2 '19 at 16:29
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    $\begingroup$ Your $\gamma$ is the same as the $\pi$ in my picture. How do you define $\varphi$ and $G'$? $\endgroup$
    – user546106
    Apr 2 '19 at 16:30
  • $\begingroup$ I've taken the liberty of having $G'=\operatorname{Im}(\gamma)$, @user546106. It's probably best if you work on what $\phi$ is yourself. $\endgroup$
    – Shaun
    Apr 2 '19 at 16:34
  • $\begingroup$ I don't understand, what do you mean by "no"? $\text{Im}(\gamma) = \gamma (G) = G/N$ so you are saying $G/N \cong G/N$. What did I miss? $\endgroup$
    – William
    Jan 5 at 11:29
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Usually, the first isomorphism theorem is easily satisfied by the projection map $\pi:G\to G/N$ defined by $$ \pi (g) = gN $$ for example see this question. Then in your specific dihedral group example, you can explicitly calculate $\pi(G)$ and get that $$G/N = G/\ker(\pi)\cong\pi(G) $$

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  • $\begingroup$ I think your answer and Shaun's just restate the first isomorphism theorem. $\endgroup$
    – user546106
    Apr 2 '19 at 16:25
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    $\begingroup$ Well to answer your question "How to find out what $G/N$ is isomorphic to?", it is equivalent to answer "how do you compute the right hand side of the first isomorphism theorem?". And we both suggest that you can compute the image of $G$ under the projection map to answer your question. Is this clearer? $\endgroup$
    – NazimJ
    Apr 2 '19 at 16:30

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