Let the CW complex \(X\) be obtained from \(S^1 \vee S^n\), \(n \geq 2\), by attaching a cell \(e^{n+1}\) by a map representing the polynomial \(p(t) \in \mathbb{Z}[t, t^{-1}] \approx \pi_n(S^1 \vee S^n)\), so \(\pi_n(X) \approx \mathbb{Z}[t, t^{-1}] / \bigl(p(t)\bigr)\). Show \(\pi_n'(X)\) is cyclic and compute its order in terms of \(p(t)\). Give examples showing that the group \(\pi_n(X)\) can be finitely generated or not, independently of whether \(\pi_n'(X)\) is finite or infinite.
Solution
Let \[ R=\mathbb Z[t,t^{-1}],\qquad M=\pi_n(X)\cong R/(p(t)). \] The quotient \(\pi_n'(X)\) is obtained by imposing the relations \[ t\alpha=\alpha \] for all \(\alpha\in M\). Hence \[ \pi_n'(X)=M/(t-1)M \cong R/(p(t),t-1). \] Evaluation at \(t=1\) gives \[ R/(t-1)\cong\mathbb Z, \] so \[ \boxed{\pi_n'(X)\cong\mathbb Z/(p(1)).} \] Thus it is cyclic of order \(|p(1)|\) when \(p(1)\ne0\), and infinite cyclic when \(p(1)=0\).
The four finiteness possibilities all occur.
- \(p(t)=t+1\): then \(M\cong\mathbb Z\), while \(\pi_n'(X)\cong\mathbb Z/2\).
- \(p(t)=t-1\): then \(M\cong\mathbb Z\), while \(\pi_n'(X)\cong\mathbb Z\).
- \(p(t)=2t-1\): then \(M\cong\mathbb Z[1/2]\), not finitely generated as an abelian group, while \(p(1)=1\) so \(\pi_n'(X)=0\).
- \(p(t)=0\): then \(M=\mathbb Z[t,t^{-1}]\), not finitely generated as an abelian group, while \(\pi_n'(X)\cong\mathbb Z\).
Hence finite generation of \(\pi_n(X)\) and finiteness of \(\pi_n'(X)\) are independent.