Stability group of subnormal series of finite p-group is p-group

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Statement

Let P be a finite p-group, i.e., a group of prime power order where the underlying prime is p. Suppose there is a subnormal series:

\{ e \} \triangleleft P_1 \triangleleft P_2 \triangleleft \dots \triangleleft P_n = P

Consider the subgroup of \operatorname{Aut}(P) given as the group of stability automorphisms of this subnormal series, i.e., as follows:

K := \{ \sigma \in \operatorname{Aut}(P) \mid \sigma(gP_i) = gP_i \ \forall 0 \le i \le n - 1, g \in P_{i+1} \}

Then K is a p-group.

Related facts

Converse

A converse of sorts is true for this statement (note that the converse actually establishes a normal series, and not just a subnormal series):

Generalizations

Other related facts about groups of prime power order

Facts used

  1. Cauchy's theorem: This states that given a finite group, and a prime q dividing the order of the group, there is an element of the finite group of order exactly q.

Proof

Proof idea

By induction, the proof can be reduced to the case where we have a subnormal series of length two, and an automorphism that fixes both the normal subgroup and the quotient. Next, we iterate this automorphism by its order, and show that the order of the automorphism must equal the order of some element in the normal subgroup, which is hence a prime power.

Proof details

Given: P is a finite p-group, i.e., a group of prime power order where the underlying prime is p. Suppose there is a subnormal series:

\{ e \} \triangleleft P_1 \triangleleft P_2 \triangleleft \dots \triangleleft P_n = P

Consider the subgroup of \operatorname{Aut}(P) given as the group of stability automorphisms of this subnormal series, i.e., as follows:

K := \{ \sigma \in \operatorname{Aut}(P) \mid \sigma(gP_i) = gP_i \ \forall 0 \le i \le n - 1, g \in P_{i+1} \}

To prove: K is also a p-group

Proof: By fact (1), we know that if K were not a p-group, we can find elements in it whose order is equal to one of the other prime factors of K.

Thus, it suffices to show the following: For any element \sigma \in K such that \sigma has order m, relatively prime to p, \sigma to be the identity automorphism.

We prove the claim by induction on the order of P.

\sigma is a stability automorphism for the subnormal series of P. Truncating this series at P_{n-1}, we get a subnormal series for P_{n-1}, and \sigma gives a stability automorphism for that subnormal series as well. Applying the induction assumption to P_{n-1}, \sigma gives a stability automorphism of order relatively prime to p on P_{n-1}, so \sigma must act as the identity on P_{n-1}. Also, by our assumption, \sigma acts as the identity on the quotient P/P_{n-1}.

Thus, we need to show that if \sigma acts as identity on P_{n-1} and on P/P_{n-1}, and has order m relatively prime to p, then \sigma is the identity automorphism.

Pick g \in P. Then, consider the element \sigma(g). Clearly, we can write \sigma(g) = gh for some h \in P_{n-1}, because \sigma preserves the cosets of P_{n-1}. Further, since \sigma(h) = h, we have, by induction, that for every positive integer r:

\sigma^r(g) = gh^r

In particular, we have:

g = \sigma^m(g) = gh^m \implies e = h^m

But h \in P_{n-1} has order a power of p, and m is relatively prime to p, forcing h = e. So \sigma(g) = g. This argument works for every g \in P, so every element is fixed by the action of \sigma so \sigma is the identity automorphism.

References

Textbook references