Does anyone know a good place to find an example of a group, $G$, of "small" order in particular finite such that there are subgroups $H,K$ of $G$ with $HK=G$ and $G$ a Zappa-Szep product of $H$ and $K$ (in particular neither $H$ nor $K$ is normal?
Small example of Zappa-Szep product of finite groups
5
$\begingroup$
abstract-algebra
group-theory
finite-groups
-
1An easy example is $G=S_4$ with $H=\langle (1,2,3),(1,2) \rangle$ and $K=\langle (1,2,3,4) \rangle$, but there are examples of order $16$ as shown in Jim Belk's answer. – 2017-02-23
1 Answers
4
The smallest nontrivial examples have order $16$. One of them is $$ \mathbb{Z}_2 \times D_8 \,=\, \langle a, r,s \mid a^2=r^4=s^2=[a,r]=[a,s]=1,sr=r^{-1}s\rangle, $$ which is the Zappa-Szép product of the subgroups $\{1,a,s,as\}$ and $\{1,ar^2,rs,ar^3s\}$.
Other examples include $S_4$ (as Derek Holt mentions in the comments) and $D_{24}$, which is the Zappa-Szép product of the subgroups $\langle r^4,s\rangle \cong D_6$ and $\langle r^6,rs\rangle \cong V$.
-
0I found one. $A_5$ is a simple group and has 60 elements. Since $A_4$ and $\mathbb{Z}_5$ can be embedded in $A_5$ and they intersect trivially and have 12 and 5 elements respectively, we have that $A_5=A_4\mathbb{Z}_5$. – 2017-02-23