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.

  • 1
    $\begingroup$ It will help to know the classification of groups of order $4$. $\endgroup$ – Angina Seng May 7 '19 at 4:18
  • 1
    $\begingroup$ You haven't yet answered the question though, which is to find how many non-isomorphic groups of order $76$ there are and to describe their isomorphism types. Your first case gives rise to the two abelian groups of that order: $C_2 \times C_{38}$ and $C_{76}$. What about the other cases? $\endgroup$ – the_fox May 7 '19 at 4:32

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 $76$.

  • $\begingroup$ Thanks a lot for the answer. Although, I'm still confused about the second case in your answer. In particular, could you please elaborate on why the group $C_{38}\rtimes_\psi C_2$? I thought it should be $C_{19}\rtimes_\psi V$? $\endgroup$ – msd15213 May 8 '19 at 4:11
  • $\begingroup$ The two groups you mentioned are isomorphic. One way to think about it is that because $\psi$ is trivial on one $C_2$ factor, when we restrict our attention to that factor, the semidirect product is a direct product. $\endgroup$ – Rylee Lyman May 8 '19 at 10:39

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