Groups of Order 168 (6.2.26)

Dummit and Foote Abstract Algebra, section 6.2, exercise 26:

Evaluate the validity of the following statement: $$|G|=168 \land~n_7(G)>1 ⇒ G~\text{is simple}$$ Proof: The statement is false. Let $Z_2^3 \rtimes Z_7$ denote the semidirect product afforded by a homomorphism of the generator of $Z_7$ into a generator of a cyclic Sylow 7-subgroup of $Aut(Z_2^3)≅GL_3(Z_2)$ of order $168=2^3*3*7$. This implies $Z_7 \not \trianglelefteq Z_2^3 \rtimes Z_7$, so that $n_7(Z_7 \rtimes Z_2^3)>1$. Now, observe the group $G=(Z_2^3 \rtimes Z_7) \times Z_3$ of order 168. We have the former group as a subgroup of this one, so that $n_7(G)>1$. Furthermore, if $((a,b),c)∈P_3∈Syl_3(G)$, then $((a,b),c)^3=((a,b)^3,c^3)=((1,1),1)$, and since $(a,b)$ is an element of the subgroup of order $56$, we must have $(a,b)=(1,1)$ so that there are only $|Z_3|=3$ distinct elements of $G$ satisfying $x^3=1$, therefore $n_3(G)=1$, $P_3 \trianglelefteq G$ and $G$ is not simple.$~\square$