Free $p$-group of nilpotency class $4$

Question

I was reading through the article Words and characters in finite $p$-groups, and I got stuck on their presentation of the free $p$-group of nilpotency class $4$ as $G = 1 + J$ where $J = I/I^4$ and $I$ is the ideal generated by $X, Y$ in the algebra of polynomials in the non-commuting indeterminates $X, Y$ with coefficents from $\mathbb{F}_p$.

No matter how I think about that construction, I can’t understand how $G$ is supposed to be a group, and how $1+J$ is even defined (as $I / I^4$ is an independent ring).

Thank you for your help.

Answer 1

Let $R$ be a ring and let $I$ be a two-sided ideal of $R$. Given an exponent $n ≥ 1$, we can consider the ideal $I^n$ of $R$ and form the quotient ring $R / I^n$. In this quotient ring, we have the two-sided ideal $I / I^n$. All elements of $I / I^n$ are nilpotent in $R / I^n$.

Let us recall the following observation about nilpotent elements in a ring:

Observation (geometric series). Let $S$ be a ring and let $x$ be a nilpotent element of $S$. The element $1 - x$ is invertible in $S$, with inverse given by $$ (1 - x)^{-1} = 1 + x + x^2 + x^3 + \dotsb = \sum_{n = 0}^∞ x^n \,. $$ Similarly, the element $1 + s$ is invertible with inverse given by $$ (1 + x)^{-1} = 1 - x + x^2 - x^3 + \dotsb = \sum_{n = 0}^∞ (-1)^n x^n \,. \tag{$\ast$} $$

This observation has a direct consequence:

Corollary. Let $S$ be a ring. Let $J$ be an additive subgroup of $S$ closed under multiplication (a “non-unital subring”) and consisting of nilpotent elements. The set $1 + J = \{1 + x \mid x ∈ J\}$ is a subgroup of $S^×$ (the group of invertible elements of $S$).

Proof. Any element $x$ of $J$ is nilpotent in $S$, whence $1 + x$ is invertible in $S$. In other words, $1 + J$ is contained in $S^×$. The neutral element of $S^\times$ is $1$, which is contained in $1 + J$ because $0$ is contained in $J$. For any two elements $x$ and $y$ of $J$, we have $$ (1 + x)(1 + y) = 1 + x + y + xy \,. $$ The term $x + y + xy$ is again contained in $J$, whence $(1 + x)(1 + y)$ is again contained in $1 + J$. This shows that $1 + J$ is closed under multiplication. For any element $x$ of $J$ the element $\sum_{n = 1}^∞ (-1)^n x^n$ is again contained in $J$, whence the element $(1 + x)^{-1} = 1 + \sum_{n = 1}^∞ (-1)^n x^n$ is again contained in $1 + J$. This shows that $1 + J$ is closed under taking inverses.

Corollary of the corollary. Let $S$ be a ring and let $J$ be an ideal of $S$ consisting of nilpotent elements. Then $1 + J$ is a subgroup of $S^×$.

Going back to the beginning of this post, we find that for the ideal $J ≔ I / I^n$ of $R / I^n$ the set $1 + J$ is a subgroup of $(R / I^n)^×$. In other words, $1 + J$ becomes a group via the multiplication from $R / I^n$.