Show \(\pi_3(S^1 \vee S^2)\) is not finitely generated as a module over \(\mathbb{Z}[\pi_1(S^1 \vee S^2)]\) by considering Whitehead products in the universal cover, using the results in Example 4.52. Generalize this to \(\pi_{i+j-1}(S^1 \vee S^i \vee S^j)\) for \(i, j > 1\).
Solution
Let \[ X=S^1\vee S^2. \] Its universal cover is an infinite line with a copy \(S_k^2\) of \(S^2\) attached at each integer \(k\in\mathbb Z\). Let \[ \iota_k\in\pi_2(\widetilde X) \] be the inclusion of the \(k\)-th sphere. The deck generator \(t\) acts by \[ t\iota_k=\iota_{k+1}. \]
Example 4.52 and the Hilton–Milnor calculation show that in \[ \pi_3(\widetilde X) \] the Whitehead products \[ [\iota_a,\iota_b],\qquad a<b, \] form independent infinite-order summands, in addition to the individual \(\pi_3(S_k^2)\)-summands. Translation acts by \[ t[\iota_a,\iota_b]=[\iota_{a+1},\iota_{b+1}]. \] Thus the difference \[ d=b-a>0 \] is invariant under the \(\mathbb Z[t,t^{-1}]\)-action. For every \(d\ge1\), the class \[ [\iota_0,\iota_d] \] lies in a different module orbit, and these classes are independent. Hence no finite set can generate all of them: \[ \boxed{\pi_3(S^1\vee S^2)\text{ is not finitely generated over }\mathbb Z[\pi_1].} \]
More generally, let \[ Y=S^1\vee S^i\vee S^j, \qquad i,j>1. \] In the universal cover there are copies \(S^i_a\) and \(S^j_b\) at every integer. If \(\alpha_a\) and \(\beta_b\) denote their fundamental homotopy classes, then \[ [\alpha_a,\beta_b] \in\pi_{i+j-1}(\widetilde Y) \] are independent Hilton–Milnor summands. Deck translation sends \[ [\alpha_a,\beta_b]\mapsto[\alpha_{a+1},\beta_{b+1}], \] so again the difference \(b-a\) labels infinitely many distinct module orbits. Therefore \[ \boxed{\pi_{i+j-1}(S^1\vee S^i\vee S^j) \text{ is not finitely generated over }\mathbb Z[\pi_1].} \]