Omega-1 of odd-order p-group is coprime automorphism-faithful
This article gives the statement, and possibly proof, of a particular subgroup of kind of subgroup in a finite group being coprime automorphism-faithful. In other words, any non-identity automorphism of of the whole group, of coprime order to the whole group, that restricts to the subgroup, restricts to a non-identity automorphism of the subgroup.
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Template:Minimal counterexample method
Contents
Statement
Suppose is an odd prime, and is a finite -group (i.e., a Group of prime power order (?)). Let be the first omega-subgroup of , i.e.:
.
Then, is a Coprime automorphism-faithful subgroup (?) of : if is a non-identity automorphism of of order relatively prime to , then the restriction of to is also a non-identity automorphism.
Related facts
Failure for the prime two
For the prime two, the quaternion group is a group where is not coprime automorphism-faithful: for instance, any non-identity inner automorphism has order two, but acts as the identity on , which is equal to the center.
However, a partial equivalent is true at the prime two:
Automorphism of finite 2-group of Mersenne prime order acts nontrivially on Omega-1
Applications
- Odd-order p-group has coprime automorphism-faithful characteristic class two subgroup of prime exponent: This is obtained by combining the given statement with Thompson's critical subgroup theorem
Facts used
- Structure lemma for p-group with coprime automorphism group having automorphism trivial on invariant subgroups
- Frattini-in-center odd-order p-group implies p-power map is endomorphism
- Stability group of subnormal series of p-group is p-group
Proof
Given: An odd prime number , a finite -group , an automorphism of such that the order of is relatively prime to , and such that the restriction of to is the identity map.
To prove: is the identity map on .
Proof: We prove the claim by induction on the order of the group, so we can assume that the result is true for every proper subgroup of . Also, we assume that is a non-identity automorphism and derive a contradiction.
First, note that for , . Since acts as the identity on , we conclude that if is -invariant, acts as the identity on . Now, if is a non-identity automorphism, apply fact (1) for the element and the cyclic group generated by , acting on . The upshot:
is either elementary Abelian or special. If is Abelian, and acts nontrivially on .
The case that is elementary Abelian can be eliminated, because in that case , so being the identity on forces it to be the identity on . Thus, the only case left is that is a special group.
By fact (2), the -power map is an endomorphism, so for any , . Now, since is elementary Abelian. Further, since is special itself is elementary Abelian, so , and consequently acts as the identity on . Thus, , so .
Thus, for any , and are in the same coset of . Thus, acts as the identity on . By assumption, also acts as the identity on . Combining this with fact (3), we obtain that is the identity map.
References
Textbook references
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 184, Theorem 3.10, Section 5.3 (-automorphisms of -groups)