So I've been working on this problem for my qual prep class, and I have it all down except for one detail. I'm doing it by semidirect products, and with the Sylow $p$ group normal, choosing the case when the Sylow 2 subgroup is cyclic.
It's in the special case when the prime is congruent to $1$, mod $4$. Then we get an extra subgroup, as there's an extra automorphism of order 4. My problem is....how do I find this automorphism? I'd know it'd be the generator of the group to the $\frac {p-1} 4$ power, but I was looking online and there didn't seem to be a general formula for finding a generator of a multiplicative group of the finite field of order $p$. So...is there another way of specifying this automorphism of order 4 in the group of automorphisms of Cyclic groups of prime order who are congruent to 1 mod 4?
Or am I stuck just saying "Pick that element!"
If $p \equiv 1$ mod $4$, $p=1+4m$ then $(p-1)! \equiv -1$ mod $4$, (Wilson). Hence $(2m)!\prod\limits_{i=0}^{2m-1+}(2m+1+i) \equiv -1$, now $4m+1 \equiv 0$ so $2m+1 \equiv -2m$, hence $(2m)!\prod\limits_{i=0}^{2m-1+}(-2m+i) \equiv -1$, so $(2m)!\prod\limits_{i=0}^{2m-1+}(-1)(2m-i) \equiv -1$ therefore $(2m)!(-1)^{2m}\prod\limits_{i=0}^{2m-1}(2m-i) \equiv -1$, hence $[(2m)!]^{2} \equiv -1$, so this tells you that the class $m$ of $(\frac{p-1}{2})!$ has order $4$. Now, multiplication by $m$ is an automorphism of order 4. $(-m)$ is the other element of order 4, and both are the unique ones, as $x^{4}=1$ has $4$ solutions, two of which are $1$ and $-1$.