How to Understand Linear Recurrence Relations in Group Automorphism Towers

Hello! This is a fascinating problem in group theory that explores the relationship between groups and their automorphism groups. Let me help you understand what's being asked and how to approach it.

What We're Solving:

We want to find values of $r$ such that there exists a finite group $G$ where the automorphism tower (starting with $G_0 = G$, then $G_1 = \text{Aut}(G)$, $G_2 = \text{Aut}(\text{Aut}(G))$, etc.) satisfies $|G_{i+1}| = r|G_i|$ for all $i$. We're looking for constants $r$ that make this linear recurrence work.

The Approach:

The key insight is to examine specific families of groups where we can compute automorphism groups explicitly. We need to find examples where the ratio $|\text{Aut}(G)|/|G|$ is constant throughout the tower. This requires understanding both the structure of groups and their automorphism groups.

Step-by-Step Solution:

Step 1: Understand what we need For the recurrence $|G_{i+1}| = r|G_i|$ to hold, we need $|\text{Aut}(G_i)|/|G_i| = r$ for all $i$ in our tower.

Step 2: Consider simple abelian groups Let's examine $G = \mathbb{Z}/p\mathbb{Z}$ where $p$ is prime.

• $|G| = p$
• $\text{Aut}(G) \cong (\mathbb{Z}/p\mathbb{Z})^*$ (the multiplicative group of units)
• $|\text{Aut}(G)| = p-1$
• So $r = \frac{p-1}{p}$

Step 3: Check if this works for a tower For $p = 2$: $G_0 = \mathbb{Z}/2\mathbb{Z}$, so $G_1 = \text{Aut}(\mathbb{Z}/2\mathbb{Z}) = \{1\}$ (trivial group) The ratio is $(2-1)/2 = 1/2$, but then the tower terminates.

Step 4: Look for more promising examples Consider elementary abelian $p$-groups: $G = (\mathbb{Z}/p\mathbb{Z})^n$

• $|G| = p^n$
• $\text{Aut}(G) \cong GL_n(\mathbb{F}_p)$ (invertible $n \times n$ matrices over $\mathbb{F}_p$)
• $|\text{Aut}(G)| = (p^n - 1)(p^n - p)(p^n - p^2)\cdots(p^n - p^{n-1})$

Step 5: Find a working example For $G = (\mathbb{Z}/2\mathbb{Z})^2$:

• $|G| = 4$
• $\text{Aut}(G) \cong GL_2(\mathbb{F}_2) \cong S_3$ (symmetric group on 3 elements)
• $|\text{Aut}(G)| = 6$
• So $r = 6/4 = 3/2$

The Answer:

One concrete value that works is $r = 3/2$, achieved by starting with $G = (\mathbb{Z}/2\mathbb{Z})^2$. However, verifying that this ratio persists throughout the entire automorphism tower requires checking that $\text{Aut}(S_3)$ and subsequent groups maintain this ratio, which becomes quite complex.

The general question remains largely open in research, with only scattered results known for specific families of groups. Values like $r = 1$ (complete groups), $r = 3/2$, and certain other rational numbers have been shown to work, but a complete characterization is unknown.

Memory Tip:

Think of this as asking: "When does the 'complexity growth rate' (measured by group order) stay constant as we climb the automorphism tower?" The answer depends on finding groups with very regular automorphism structure - a rare and beautiful phenomenon in group theory!

This is an active area of research, so don't be discouraged if the complete answer seems elusive - mathematicians are still working on it too!