# 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.

• It will help to know the classification of groups of order $4$. – Angina Seng May 7 '19 at 4:18
• 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? – 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 $$76$$.
• 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 $C_{19}\rtimes_\psi V$? – msd15213 May 8 '19 at 4:11
• 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. – Rylee Lyman May 8 '19 at 10:39