# 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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## Statement

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:

## 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.