Determine all groups of order $76$ upto isomorphism Determine all groups of order $76 = 2^2 \cdot 19$ upto isomorphism.
My solution
The Sylow-19 subgroup is normal and cyclic. Call it $P_{19}$ and suppose $P_{19} = \langle z\rangle$. Let a Sylow-2 subgroup be given and call it $P_2$. Suppose $P_2 = \{1, a, b, c\}$.
It is easy to check that $G\cong P_{19}\rtimes_\phi P_2$, where $\phi: P_2\to\text{Aut}(P_{19})$ is a homomorphism.
For any $x\in P_2$, $\phi(x)$ can have order either $1$ or $2$, which gives four possibilities for $G$ (upto isomorphism).
Note that $\phi(1) = 1$, the trivial automorphism on $P_{19}$.
case 1:
$\phi(a) = \phi(b) = \phi(c) = 1$, which implies $G\cong P_{19}\times P_2$.
case 2:
$\phi(a) = 2,\,\phi(b) = \phi(c) = 1$. Then $\phi(a)$ maps $z$ to $z^{10}$.
case 3:
$\phi(a) = \phi(b) = 2,\,\phi(c) = 1$.
case 4:
$\phi(a) = \phi(b) = \phi(c) = 2$.
One can write out the group table for all four cases explicitly. This completely describes all possible groups of order $76$ upto isomorphism.
My question
Is my solution correct? I have a feeling that I might have missed something. For example, that $P_2$ has order $4$ may give more information on $a, b, c$ (e.g., they commute with each other), which may make one or more of the four cases impossible.
Any help would be greatly appreciated.
 A: Case 4 does not occur. There are two groups of order $4$: $C_4$ and $V = C_2 \times C_2$. 
Since $C_4$ is cyclic, $\phi$ is determined by where it sends the generator $t$. In particular, to yield a nonabelian group, $\phi(t)$ must have order two, which forces us into Case 3. Actually, in this case, we can say $\phi(t)$ maps $z$ to $z^{10}$. This exhausts the possibilities for $P_2 \cong C_4$. This gives us the group $C_{19} \rtimes_\phi C_4$.
In the case where $P_2\cong V$, we again have only one nonabelian possibility: I claim that up to an isomorphism of $V$, every map $V \to C_2$ is injective on one $C_2$ factor and kills the other. (You should check this!) This gives us the group $C_{38}\rtimes_\psi C_2$, where $\psi \colon C_2 \to \operatorname{Aut}(C_{38}) \cong \operatorname{Aut}(C_{19})\times \operatorname{Aut}(C_2)$ sends the generator $z$ of $C_{19}$ to $z^{10}$.
These groups are distinct (check the orders of elements!), as are the two groups $P_{19}\times P_2$, giving us a total of four groups of order $76$.
