Quotient group $Q$: 1) is it normal, 2) does $G=Q\times N$? I know that if we form the direct product group
$$G=G_1\times G_2$$
then both $G_1$ and $G_2$ are normal subgroups of $G$. I also know that if a group $G$ has a normal subgroup $N\triangleleft G$, then we define a quotient group
$$Q=G/N$$
where the elements are cosets.
My questions:

*

*Is $Q$ also a normal subgroup?

*Can be recover $G$ by some product of $Q$ and $N$, for example $Q\times N$?

 A: *

*$Q$ is not necessarily a normal subgroup, take the example of $SO(3)=SE(3)/T(3)$, i.e., rotation around the origin is the quotient of special Euclidean transform up to a translation. However rotating around the origin is not a normal subgroup. $t\ r \ t^{-1}$ is not a rotation around the origin in general, but a rotation around $t$.


*Sometimes it can be a direct or semidirect product, but some splitting conditions need to be satisfied.
A: To put my comments in the form of a formal answer:
The quotient group isn't even a subgroup because the elements of $Q$ are cosets ($gN,\ \forall g\in G$) which are clearly not elements of your group $G$. For a more mathy answer, please see this. Although one can sometimes find isomorphic subgroups to the quotient group, this is not always guaranteed. Let me elucidate a bit more on this. Consider $D_{3}=\{e,c,c^{2},b,bc,bc^{2}\}$ and $H=\{e,c,c^{2}\}$ which is a normal subgroup. Now, the quotient group $Q$ consists of the cosets $gH, \ \forall g\in D_{3}$ which reads
$$Q=D_{3}/H = \{eH=H,bH\}\cong C_{2},$$
i.e. it is isomorphic to $C_{2}$. For something a bit more non-trivial, consider the dicyclic group of order $3$, i.e. $Dic_{3} = \{e,x,a,a^{2},a^{3},a^{4},a^{5},x,ax,a^{2}x,a^{3}x,a^{4}x,a^{5}x\}$ and it is easy to show that it's center $Z = \{e,a^{3}\}$, and which is by definition is a normal subgroup of $Dic_{3}$. Quotienting yields
$$Q = Dic_{3}/Z = \{\{a^{3},e\},\{a,a^{4}\},\{a^{2},a^{5}\},\{x,a^{3}x\},\{ax,a^{4}x\},\{a^{2}x,a^{5}x\}\}\cong D_{3},$$
i.e. it is isomorphic to $D_{3}$, which is a subgroup of $Dic_{3}$ but not a normal subgroup of it. Extending this you can show that $Q = Dic_{n}/Z\cong D_{n}$. This finding of something that is isomorphic to your quotient is not useless since it aids us in completing the character table for higher dimensional representations via the process of lifting. with reference to the latter example, I have provided, you can define a map $\psi: Dic_{3}/Z\to D_{3}$ and convince yourself that $\psi$ is indeed an isomorphism since the inverse to $\psi$ exists (since the image in $D_{3}$ has a unique preimage and $\psi^{-1}$ preserves the group property). We can now identify that there exists a $1:1$ relation between the irreps of $Dic_{3}$ with $Z\leq Ker(\rho)$, where $\rho: Dic_{3}\to GL(n,\mathbb{C})$ and the irreps of $Dic_{3}/Z$, or put in a more succinct way,
$$D^{Dic_{3}}(g) = Dic_{3}^{\left(Dic_{3}/Z\right)}(gDic_{3}).$$
With this identification, you can simply "lift" the corresponding values of the characters in the table of $D_{3}$ to the table of $Dic_{3}$ (for example, $\chi_{Dic_{3}}(a) = \chi_{Dic_{3}}(\psi(\{a^{2},a^{5}\})) = \chi_{D_{3}}(c) = \chi_{D_{3}}(c^{2}) = -1$, and so on).
