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

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, $$|P| < |G|$$.

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!

• Please ask one question at a time. Jul 24 at 19:06
• @Shaun Sorry, I just clumped them together because they referenced the same theorems and were very closely related. I’ll create a second post with the second question and link it to this one Jul 24 at 21:20

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 $$i and $$j>n$$.