Skip to main content
edited body
Source Link
Tony
  • 111
  • 3

Let $F_n$ be a free group on $n$ generators. Fix a prime $p$. Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series $$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$ Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups. Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$. They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$.

Question : What are $H_k(N^p_n;\mathbb{Z})$$H_2(N^p_n;\mathbb{Z})$ and $H_k(N^p_n;\mathbb{F}_p)$$H_2(N^p_n;\mathbb{F}_p)$?

Let $F_n$ be a free group on $n$ generators. Fix a prime $p$. Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series $$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$ Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups. Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$. They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$.

Question : What are $H_k(N^p_n;\mathbb{Z})$ and $H_k(N^p_n;\mathbb{F}_p)$?

Let $F_n$ be a free group on $n$ generators. Fix a prime $p$. Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series $$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$ Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups. Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$. They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$.

Question : What are $H_2(N^p_n;\mathbb{Z})$ and $H_2(N^p_n;\mathbb{F}_p)$?

Source Link
Tony
  • 111
  • 3

Second homology group of free nilpotent p-group

Let $F_n$ be a free group on $n$ generators. Fix a prime $p$. Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series $$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$ Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups. Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$. They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$.

Question : What are $H_k(N^p_n;\mathbb{Z})$ and $H_k(N^p_n;\mathbb{F}_p)$?