In the paper “Elementary Proofs of Some Results of Cobordism Theory Using Steenrod Operations” by Quillen, he mentions in equation 4.3, page 44 (PDF page 16), the existence of an exact sequence$\require{AMScd}$ \begin{CD} h^q(\mathrm{pt}) @>{\emptyset(v)}>> h^q(\mathbf{B}Z_k) @>{v}>> h^{q+2}(\mathbf{B}Z_k) \end{CD}
where $h$ is some complex oriented cohomology theory, $Z_k$ is the cyclic group on $k$ letters, $v$ is the euler class of the line bundle on $\mathbf{B}Z_k$ which corresponds to the the character of the permutation relation, and $\emptyset(v)$ is the divided $k$-series, that is $\tfrac{[k]_h(v)}{v}$, with $[k]_h(v)$ the multiplication by $k$ in the formal group law of $h$.
According to him this comes from an commutative diagram \begin{CD} h^{q+1}(S^1_{hZ_k}) @>>> h^{q}(\mathbf{B}Z_k) @>{v}>> h^{q+2}(\mathbf{B}Z_k) \\ @AA{\cong}A @AA{\emptyset(v)}A \\ h^{q+1}(S^1) @<{\cong}<< h^{q}(\mathrm{pt}) \end{CD}
Here the top row is the Thom-Gysin sequence, the left vertical isomorphism comes from the fact that $S^1_{hZ_k}$ is homotopy equivalent to $S^1/Z_k$, as the action of $Z_k$ is free, which in turn is itself homtopy equivalent to the $S^1$, and the bottom horizontal isomorphism is the suspension isomorphism.
My Question is: Why does the square commute?
Quillen states that it is possible to show the commutation of this square, but as this is not in the spirit of the paper, he will not elaborate on it further. What he did until this point was to show that $j:Q\to B$ is a $Z_k$ torsor, as is $p:\ast\to\mathbf{B}Z_k$, then $j_\ast1=\emptyset(e(L))\in h^\ast(B)$, where $L$ is the line bundle associated to this torsor. So I guess that the right vertical morphism is $p_\ast$.
Any hint or reference on why this diagram commutes would be really appreciated.