How do I find the groups D10/N where N is the normal subgroups to D10?
I know that the definition is aN for all a $\in$ D10. But I am unsure what the group, for example, D10/ D10 looks like? Any help would be much appreciated.
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$\begingroup$The standard reference (on this site at least) for dihedral groups are the notes by K. Conrad. Theorem $2.3$ and its proof answers the question. All quotients $D_{10}/N$ are again dihedral groups. All normal subgroups of $D_{10}$ are listed in Theorem $3.8$.
$\endgroup$ $\begingroup$Summarizing the comments:
In general, a quotient group $G/N$ is a group consisting of the cosets of $N$, under the multiplication rule $(aH)(bH) = abH$. This is a valid group if and only if $N$ is a normal subgroup of $G$.
In your case, $G = D_{10}$ is the dihedral group with $10$ elements. As you noted, it has three normal subgroups, namely $\{e\}$, $G$, and $\langle s\rangle$, where the latter is the rotation subgroup of order $5$. Now, $G$ and $\{e\}$ are always normal in any group. And in this particular group, $\langle s \rangle$ is normal because it has index $2$.
Let us examine each of the quotient groups.
- If $N = G$ then $G/N = G/G$. This group has order $|G| / |N| = |G| / |G| = 10 / 10 = 1$, i.e., it is the trivial group. It contains one element, which is the coset $G$ itself. We may write $G/G = \{G\}$ if we want to list the elements explicitly.
- If $N = \{e\}$, then $G/N = G/\{e\}$. This group has order $|G| / |\{e\}| = 10 / 1 = 10$. So the group contains $10$ elements. They are the cosets of $\{e\}$, each of which is of the form $g\{e\}$ for some $g \in G$. Moreover, distinct elements $g,h \in G$ give us distinct cosets $g\{e\}$ and $h\{e\}$. There is an obvious isomorphism between this group of cosets and the original group itself: just map $g \in G$ to $g\{e\} \in G/\{e\}$. So, in this case, $G/N = G/\{e\}$ is isomorphic to $G$ itself.
- If $N = \langle s\rangle$ then $G/N = G/\langle s\rangle$. This group has order $|G| / |\langle s\rangle| = 10 / 5 = 2$. So the group contains $2$ elements. They are the cosets of $\langle s \rangle$. One of these cosets is $\langle s \rangle$ itself. The other can be expressed as $g \langle s \rangle$ where $g$ is any element of $G$ which is not in $\langle s \rangle$. Note that any group of order $2$ (or more generally, any prime number) is cyclic, so there is only one group (up to isomoprhism) of this order. Therefore $G/N$ is cyclic of order $2$.
One more thing to emphasize: as mentioned in the comments, a quotient group $G/N$ is not a subgroup of $G$. This is a common misconception when first learning about quotient groups. Rather, $G/N$ is a separate group, formed from the cosets of $N$. You can think of it as looking at the group $G$ through eyeglasses which "blur" your vision in such a way that $N$ looks like a single "element" instead of being composed of individual elements.
In some cases, $G$ will contain a subgroup which is isomoprhic to $G/N$, but in general this is not the case. Indeed, in general, $G$ need not even contain a subgroup with the same order as $G/N$.
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