Seriously, it's just a continuation of Relating the automorphisms of $P$ and those of $P/Φ(P)$, an exercise in Section 6.1 of Abstract Algebra: Dummit and Foote. A lot of typing:
I assume I had part (e) done and now I'm going to do part (f).
26. (f) Let $p$ be a prime. Let $P$ be a finite $p$-group. Use part (e) and (c) to deduce that every nontrivial automorphism of $P$ of order prime to $p$ induces a nontrivial automorphism on $P/Φ(P)$. Deduce that any group of automorphisms of $P$ which has order prime to $p$ is isomorphic to a subgroup of $\operatorname{Aut}(P/Φ(P))=GL_r(\Bbb F_p)$.
where $P/Φ(P)$ is just elementary abelian of order $p^r$.
(e) Let $\sigma$ be any automorphism of $P$ of prime order $q$ with $q\neq p$. Show that if $\sigma$ fixes the coset $xΦ(P)$ then $\sigma$ fixes some element of this coset.
(c) If $x_1Φ(P),x_2Φ(P),\dots,x_rΦ(P)$ are any basis for the $r$-dimensional vector space $P/Φ(P)$ over $\Bbb F_p$ and if $x_i$ is any element of the coset $x_iΦ(P)$, then $P=\langle x_1,x_2,\dots,x_r\rangle$...
At first, I had no idea what is being 'prime to $p$'. I assume it means 'relatively prime to $p$', instead of 'a prime relatively prime to $p$', which is same as 'a prime different from $p$'. Now down to business: I know how an automorphism $\sigma$ of $P$ induces an automorphism of $P/Φ(P)$: $$\sigma_{ind}:P/Φ(P)\rightarrow P/Φ(P)$$$$\sigma_{ind}(gΦ(P))=\sigma(g)Φ(P)$$ for all $g\in P$. $\sigma_{ind}$ is said to be induced by $\sigma$. Now the contrapositive of the first statement of (f):
Suppose $\sigma_{ind}$ is trivial. I'm gonna show that either $\sigma$ is trivial, or $p\mid|\sigma|$.
Suppose by contradiction that $\sigma\neq1$ and $p\nmid|\sigma|$. Then we write $|\sigma|$ as product of primes. Suppose $|\sigma|$ is just a prime $q$ with $q\neq p$. Since $\sigma_{ind}$ fixes every element of $P/Φ(P)$, i.e., $gΦ(P)=\sigma(g)Φ(P)$ for all $g$, $\sigma$ fixes every coset of $Φ(P)$, as a result of $Φ(P)$ being a characteristic subgroup. So by (e), every coset of $Φ(P)$ contains a fixed point of $\sigma$. Among those cosets, we can find a basis for $P/Φ(P)$. By (c), we can choose an arbitrary representative from each of the cosets that form a basis and have the representatives generate $P$. So we might as well just choose those fixed points! So $P$ is generated by fixed points of $\sigma$! Contradicts to our assumption that $\sigma\neq1$.
Even if $|\sigma|$ is not a prime, that's fine. (I hope so) Every prime factor of $|\sigma|$ is not $p$. Choose one, say, $q$, and there must be some power of $\sigma$ of order $q$ (say $\sigma^{|\sigma|/q}$). i.e. $\sigma^{|\sigma|/q}$ is an automorphism of $P$ of order $q$ with $q\neq p$. Since $\sigma$ fixes every coset of $Φ(P)$, so does $\sigma^{|\sigma|/q}$. By similar arguments, $P$ is generated by fixed points of $\sigma^{|\sigma|/q}$, so $\sigma^{|\sigma|/q}=1$. So $|\sigma|\mid(|\sigma|/q)$, which is absurd.
Questions:
Is this proof correct? I know it appears lengthy and I shouldn't have separated it into two cases when $|\sigma|$ is prime and when it's not, but is it correct?
I have no idea about the second statement of (f). Maybe if $\{\sigma_1,\dots\sigma_j\}$ is a group of automorphism of $P$, then it is isomorphic to $\{\sigma_{1_{ind}}\dots\sigma_{j_{ind}}\}$? Can somebody give me a hint at least?