= Solution
Choose a representative <Sylow subgroup> with generators
$$
r=(1\ 2\ 3\ 4),\qquad s=(1\ 3),\qquad t=(5\ 6).
$$
On the first four points, $r^4=s^2=1$ and $srs=r^{-1}$, so $\langle r,s\rangle$ is the <dihedral group> of order eight. The disjoint <transposition> $t$ commutes with it and generates an independent factor of order two. Thus $G=\langle r,s,t\rangle\cong D_8\times C_2$ has order 16. Since the power of two dividing $7!$ is $2^{\lfloor7/2\rfloor+\lfloor7/4\rfloor}=2^4$, this is a Sylow 2-subgroup of $S_7$. Any other choice is conjugate to it and has conjugate series.
The dihedral relation gives $[r,s]=r^2$, while $t$ commutes with both generators. Modulo $\langle r^2\rangle$ all three generators commute, so the <commutator subgroup> is exactly $\langle r^2\rangle$, and $r^2$ is central. Hence the <lower central series> is
$$
\boxed{\Gamma_1(G)=G,\qquad \Gamma_2(G)=\langle(1\ 3)(2\ 4)\rangle,\qquad \Gamma_3(G)=1.}
$$
Within the dihedral factor only $1,r^2$ commute with both $r$ and $s$. The center of $G$ is therefore $\langle r^2,t\rangle\cong C_2\times C_2$. Its quotient has order four and is abelian, so the <upper central series> is
$$
\boxed{Z_0(G)=1,\qquad Z_1(G)=\langle(1\ 3)(2\ 4),(5\ 6)\rangle,\qquad Z_2(G)=G.}
$$
Both series show that the <nilpotency class> is two.
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