Size of $p$-subgroups of $\operatorname{Aut}(G)$, where $p$ divides the order of $G$

Question

Let $G$ be a nontrivial finite group. In his book "Finite Group Theory", M. Isaacs proves the following two results:

Corollary 3.3: Let $\sigma \in \operatorname{Aut}(G)$. Then, $o(\sigma) < |G|$.

Corollary 3.4: Let $P$ be an abelian $p$-subgroup of $\operatorname{Aut}(G)$ and suppose $p \nmid |G|$ . Then, $|P| < |G|$.

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This lead me to the following question:

Question: is the same result (3.4) true when $p \mid |G|$?

I'm guessing "no", but I know very few concrete examples of automorphism groups of finite groups. If $G$ is cyclic, for instance, $|\operatorname{Aut}(G)| = \phi(n) < n$, meaning the question is automatically true.

If $G$ is dihedral, I found $\operatorname{Aut}(G) \simeq C_n \rtimes C_n^\times$, where $|G| = 2n$ and the action is $x^p \cdot x^q = x^{pq}, x^p \in C_n^\times, x^q \in C_n$, which has order $n \phi(n)$; in particular, for cases like $n = 8$, it seems like I could find a counter-example, but I'm still struggling a bit to find abelian subgroups of this semidirect product, since I'm a bit new to these groups...

EDIT: As has been suggested in the comments, the second question has been moved to another post.

Any help is appreciated! Thanks in advance!

Answer 1

I will answer Question 1: the answer is no.

The automorphism group of an elementary abelian group of order $p^{2n}$ is isomorphic to the group ${\rm GL}(2n,p)$, which has an elementary abelian subgroup of order $p^{n^2/4}$ consisting of upper unitriangular matrices with nonzero entries above the main diagonal only in the upper right $n \times n$ block.

In other words, matrices $(\alpha_{ij})$ with $\alpha_{ii}=1$ for all $i$, and $\alpha_{ij}=0$ for $i \ne j$ except when $i<n$ and $j>n$.