Omega-1 of odd-order p-group is coprime automorphism-faithful

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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


Suppose p is an odd prime, and P is a finite p-group (i.e., a Group of prime power order (?)). Let \Omega_1(P) be the first omega-subgroup of P, i.e.:

\Omega_1(P) = \langle x \mid x^p = e \rangle.

Then, \Omega_1(P) is a Coprime automorphism-faithful subgroup (?) of P: if \sigma is a non-identity automorphism of P of order relatively prime to p, then the restriction of \sigma to \Omega_1(P) is also a non-identity automorphism.

Related facts

Failure for the prime two

For the prime two, the quaternion group is a group where \Omega_1 is not coprime automorphism-faithful: for instance, any non-identity inner automorphism has order two, but acts as the identity on \Omega_1, 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


Facts used

  1. Structure lemma for p-group with coprime automorphism group having automorphism trivial on invariant subgroups
  2. Frattini-in-center odd-order p-group implies p-power map is endomorphism
  3. Stability group of subnormal series of p-group is p-group


Given: An odd prime number p, a finite p-group P, an automorphism \sigma of P such that the order of \sigma is relatively prime to p, and such that the restriction of \sigma to \Omega_1(P) is the identity map.

To prove: \sigma is the identity map on P.

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 P. Also, we assume that \sigma is a non-identity automorphism and derive a contradiction.

First, note that for Q \le P, \Omega_1(Q) \le \Omega_1(P). Since \sigma acts as the identity on \Omega_1(P), we conclude that if Q is \sigma-invariant, \sigma acts as the identity on Q. Now, if \sigma is a non-identity automorphism, apply fact (1) for the element \sigma and the cyclic group generated by \sigma, acting on P. The upshot:

P is either elementary Abelian or special. If P is Abelian, and \sigma acts nontrivially on P/P'.

The case that P is elementary Abelian can be eliminated, because in that case P = \Omega_1(P), so \sigma being the identity on \Omega_1(P) forces it to be the identity on P. Thus, the only case left is that P is a special group.

By fact (2), the p-power map is an endomorphism, so for any g \in P, (g^{-1}\sigma(g))^p = g^{-p}\sigma(g^p). Now, g^p \in Z(P) since P/Z(P) is elementary Abelian. Further, since P is special Z(P) itself is elementary Abelian, so Z(P) \le \Omega_1(P), and consequently \sigma acts as the identity on Z(P). Thus, g^p = \sigma(g^p), so (g^{-1}\sigma(g))^p = e.

Thus, for any g \in P, g and \sigma(g) are in the same coset of \Omega_1(P). Thus, \sigma acts as the identity on P/\Omega_1(P). By assumption, \sigma also acts as the identity on \Omega_1(P). Combining this with fact (3), we obtain that \sigma is the identity map.


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