# Thompson's replacement theorem for abelian subgroups

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## Statement

Suppose $P$ is a group of prime power order (we'll call the prime $p$).

Let $\mathcal{A}(P)$ denote the set of all abelian subgroups of maximum order in $P$ (i.e., $|A| \ge |B|$ for all abelian subgroups $B$ of $P$).

Suppose $A \in \mathcal{A}(P)$ and $B$ is an abelian subgroup such that $A$ normalizes $B$ but $B$ does not normalize $A$. Then, there exists an abelian subgroup $A^*$ of $AB$ such that:

1. $A^* \in \mathcal{A}(P)$.
2. $A \cap B$ is a proper subgroup of $A^* \cap B$.
3. $A^*$ is contained in $\langle A^B \rangle = \langle A^{AB} \rangle$, i.e., it is contained in the normal closure of $A$ in $AB$. In particular, it is contained in the normal closure $\langle A^P \rangle$ of $A$ in $P$.
4. $A^*$ normalizes $A$.

We have the following additional conclusions:

1. $A^*$ normalizes $B$: Note that since $A^*$ is contained in $AB$, and $AB \le N_P(B)$ by assumption, we see that $A^*$ normalizes $B$ as well.
2. $A^* \cap B$ is a proper subgroup of $B$: This is because $A^*$ cannot contain $B$, since $A^*$ normalizes $A$ but $B$ does not normalize $A$.

Thus, all the conditions assumed for $A$ also hold for $A^*$, except possibly the fact that $B$ does not normalize $A$. Hence, the term replacement.

## Related facts

### Similar replacement theorems

For a complete list of replacement theorems, refer:

## Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A finite $p$-group $P$. $\mathcal{A}(P)$ is the set of abelian subgroups of $P$ of maximum order. $A \in \mathcal{A}(P)$, and $B$ is a subgroup of $A$ such that $A$ normalizes $B$ but $B$ does not normalize $A$.

To prove: There exists $A^* \in \mathcal{A}(P)$ such that $A \cap B$ is a proper subgroup of $A^* \cap B$, $A^*$ normalizes $A$ and $B$, and $A^*$ is contained in $AB$.

Proof: We let $C = AB$ and $N = N_B(A)$.

basic diagram of key initial constructions (steps 1-6)
More detailed diagram
Step no. Assertion Facts used Given data used Proof steps used Explanation
1 $C = AB$ is a subgroup of $P$, and $B$ is normal in $C$ -- $A$ normalizes $B$ -- [SHOW MORE]
2 $B$ normalizes $N$ -- $B$ is abelian -- [SHOW MORE]
3 $A$ normalizes $N$ -- $A$ normalizes $B$ -- [SHOW MORE]
4 $N = N_B(A)$ is normal in $C$ -- -- steps (2), (3) [SHOW MORE]
5 $N$ is a proper subgroup of $B$ -- $B$ does not normalize $A$ [SHOW MORE]
6 $B/N$ is a nontrivial normal subgroup of $C/N$ fact (3) -- steps (1), (4), (5) [SHOW MORE]
7 $B/N \cap Z(C/N)$ is nontrivial fact (2) -- step (6) Direct combination of the facts
8 Let $x \in P$ be such that its image mod $N$ is a nontrivial element in $B/N \cap Z(C/N)$. Then $M = [x,A]$ is contained in $N$ and hence in $B$ step (7) Note that step (7) guarantees the existence of such $x$. [SHOW MORE]
9 $M = [x,A]$ is an abelian subgroup of $P$ $B$ is abelian step (8) [SHOW MORE]
10 $A^* = MC_A(M)$ is an abelian subgroup of maximum order in $P$ fact (1) $A$ is abelian of maximum order step (9) Piece together.
11 $A \cap B$ is contained in $C_A(M)$ $B$ is abelian step (8) [SHOW MORE]
12 $A \cap B$ is contained in $A^* \cap B$ -- step (11) [SHOW MORE]
13 $A^*$ normalizes $A$ step (8) [SHOW MORE]
14 $M$ is not contained in $A$ step (8) [SHOW MORE]
15 $A^* \cap B$ is not contained in $A \cap B$ steps (8), (14) [SHOW MORE]
16 $A^*$ is contained in $C = AB$ steps (1), (8), (10) [SHOW MORE]
17 $A^*$ is contained in $\langle A^C \rangle$. steps (8), (9), (10) [SHOW MORE]
18 $A^*$ normalizes $B$ step (1), (16) [SHOW MORE]

The conclusions of steps (10), (12), (13), (15), (16), (17), and (18) complete the proof.

### Conceptual summary of proof

The key idea of the proof is to use fact (1) somehow, and this requires finding a suitable $x$ such that $[x,A]$ is abelian. and the resulting subgroup $A^*$ that we get is nice enough.

Working backward from this, we see that we would like that $[x,A]$ not be a subgroup of $A$, which in turn means that we want $x$ outside $N_P(A)$. On the other hand, we do want $A^*$ to normalize $A$, which means that we want $[x,A]$ to lie inside $N_P(A)$. Finally, we would like $[x,A] \subseteq B$ so that the subsequent group constructed is in $AB$. With all these considerations, we should aim to find $x$ outside $N_B(A)$ but inside $B$, but such that $[x,A] \subseteq N_B(A)$.

This motivation now allows us to work out the proof details given above. Here is a rough outline of the proof:

• Steps (1)-(6) set things up. The fact that $B$ does not normalize $A$ is used to show that $B/N$ is a nontrivial normal subgroup of $P/N$.
• The crucial steps for the construction are (7) and (8), that involve the choice of $M$. Steps (9) and (10) now use fact (1) to get a hold on $A^*$.
• The remaining steps help us get a clearer idea of the subgroup inclusions and prove the remaining desired properties for $A^*$.

## GAP implementation

Here is a GAP implementation of the constructive approach used in the proof:

ThompsonsReplacementAbelian := function(P,A,B)
local C,N,x,M,CAM,Astar;
C := Group(Union(A,B));
N := Normalizer(B,A);
x := Filtered(Set(B),y -> (not(y in N)) and IsSubgroup(N,CommutatorSubgroup(Group(y),C)))[1];
M := Group(List(Set(A),y -> x^(-1) * y^(-1) * x * y));
CAM := Centralizer(A,M);
Astar := Group(Union(M,CAM));
return Astar;
end;;