multiplicative group. The additive group of $T\rtimes_{\circ}S$ is the direct sum of the additive groups of the left ideals $A_z$. If $\alpha$ is not trivial, then the multiplicative group of $T\rtimes_{\circ}S$ is metabelian but not abelian. Furthermore, if the left ideals $A_z$ are finite and of relatively prime orders, then the multiplicative group of $T\rtimes_{\circ}S$ also is an $A$-group.

Note that, for all $u,v\in T$ and $s,s'\in S$, $(u+v,s+s')-(u,s)=(v,s'-b(u,v))$ by the definition of the sum in the asymmetric product. Thus we have $$\begin{aligned}
\lefteqn{\lambda_{((t_1,\dots ,t_n),(s_1,\dots, s_n))}((t'_1,\dots
,t'_n),(s'_1,\dots, s'_n))}\\
&=&((t_1,\dots ,t_n),(s_1,\dots, s_n))((t'_1,\dots
,t'_n),(s'_1,\dots, s'_n))\\
&&-((t_1,\dots ,t_n),(s_1,\dots, s_n))\\
&=&((t_1,\dots ,t_n)\cdot\alpha_{(s_1,\dots,s_n)}(t'_1,\dots,
t'_n),\\
&&\; (s_1,\dots
,s_n)\cdot(s'_1,\dots, s'_n))-((t_1,\dots ,t_n),(s_1,\dots, s_n))\\
&=&((t_1+f_{s_1}^{(1,1)}\cdots f_{s_n}^{(n,1)}(t'_1),\dots
,t_n+f_{s_1}^{(1,n)}\cdots f_{s_n}^{(n,n)}(t'_n)),\\
&&\; (s_1+s'_1,\dots
,s_n+s'_n))-((t_1,\dots ,t_n),(s_1,\dots, s_n))\\
&=&((f_{s_1}^{(1,1)}\cdots f_{s_n}^{(n,1)}(t'_1),\dots
,f_{s_1}^{(1,n)}\cdots f_{s_n}^{(n,n)}(t'_n)),(s'_1,\dots
,s'_n)\\
&&\quad -b((t_1,\dots ,t_n),(f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t'_1),\dots
,f_{s_1}^{(1,n)}\cdots f_{s_n}^{(n,n)}(t'_n))))\\
&=&((f_{s_1}^{(1,1)}\cdots f_{s_n}^{(n,1)}(t'_1),\dots
,f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t'_n)),\\
&&\; (s'_1-b_1(t_1,f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t'_1)),\dots ,s'_n-b_n(t_n,f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t'_n)))),\end{aligned}$$ for all $(t_1,\dots ,t_n),(t'_1,\dots ,t'_n)\in T$ and $(s_1,\dots,
s_n),(s'_1,\dots ,s'_n)\in S$. Hence it is clear that $$\lambda_{((t_1,\dots ,t_n),(s_1,\dots, s_n))}((t'_1,\dots
,t'_n),(s'_1,\dots, s'_n))\in A_z,$$ for all $((t_1,\dots
,t_n),(s_1,\dots, s_n))\in T\rtimes_{\circ}S$ and $((t'_1,\dots
,t'_n),(s'_1,\dots, s'_n))\in A_z$. Let $((t_1,\dots,
,t_n),(s_1,\dots,s_n)),((t'_1,\dots,t'_n),(s'_1,\dots,s'_n))\in
A_z$. By (\[commute\]), we have that $$\begin{aligned}
\lefteqn{((t_1,\dots,t_n),(s_1,\dots,s_n))\cdot
((t'_1,\dots,t'_n),(s'_1,\dots,s'_n))}\\
&=&((0,\dots, 0,t_z+t'_z,0,\dots,0),(0,\dots,
0,s_z+s'_z,0,\dots,0))\in A_z.\end{aligned}$$ Hence $A_z$ is a left ideal of $T\rtimes_{\circ}S$ with abelian multiplicative group.

Note that the multiplicative group 