If $A$ and $B$ are normal subgroups of group $G$, is the product of the normal subgroup $A_1$ of $A$ and the normal subgroup $B_1$ of $B$ a group?
I have tried to find the example in symmetric group($S_4$, $S_5$,$S_6$ and $S_7$) and generalized qaternion group($Q_{2^{n}} $$= \langle a, b \mid a^{2^{n-1}}=1, b^{2}=a^{2^{n-2}},$$b^{-1}ab$$=a^{-1} \rangle$). I also tried some special groups. But there does not seem to have example in group like the above. Thus, I was wondering what conditions a group $G$ needs can make sure evey product of $A_1$ and $B_1$ like the upfront is a subgroup.
We know that the product $HI$ of two subgroups $H,I$ of $G$ need not be a subgroup of $G$ if neither $H$ nor $I$ is a normal subgroup. We also know that if $A_1 \trianglelefteq A \trianglelefteq G$, then it is not necessarily true that $A_1 \trianglelefteq G$. So, it is possible that neither $A_1$, $B_1$ are normal in $G$.
Consider the Dihedral group of order $8$. $$D_8=\{e,a,a^2,a^3,x,ax,a^2x,a^3x\}$$ Two of its subgroups are $X=\{e,x, a^2, a^2x\}$ and $Y=\{e,ax,a^2,a^3x\}$ both of which are normal (and isomorphic to Klein $4$ group ($V_4$)).
Also, note $\{e,x\} \trianglelefteq X$ and $\{e,ax\} \trianglelefteq Y$ since all subgroups of $V_4$ are normal in $V_4$. However, $$\{e,x\}\{e,ax\}=\{e,x,ax,a^3\}$$ is not a subgroup of $D_8$, for $(ax)(a^3)=a^2x$ is not in the set. You can verify all of these by noting the operations of $D_8$ online; this page has a good outline of the structure of $D_8$.
So we have shown an example of the product of two normal subgroups of two other normal subgroups of $G$ that is not itself a subgroup of $G$.