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

Statement
Suppose $$p$$ is an odd prime, and $$P$$ is a finite $$p$$-group (i.e., a fact about::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 fact about::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.

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

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

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

Proof
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.

Textbook references

 * , Page 184, Theorem 3.10, Section 5.3 ($$p'$$-automorphisms of $$p$$-groups)