# How many non isomorphic groups of order 30 are there?

Let $|G|=30$. I have prove that there is the only subgroup of order $15$, which I'll denote $H$. Now I do know how to classify the group. After thinking, I made the following steps.

1) Possible order of subgroup $K$ of $G$ of order 2 are 1, 3, 5, 15.

Case 1. if $G$ contain only one element of order 2, then $G \cong Z_{30}$.

Now I cannot solve for the next steps. Please give me any hints or any other method.

• @Anindya, you really must make more efforts to write correctly your question, and not only from a grammatical point of view. For example, in (1), what does "possible subgroup...of order 2 are 1,3,5,15" mean?! Commented Feb 26, 2013 at 4:30
• I would like to point you to math.stackexchange.com/q/569226/61691. That question has been marked as a duplicate of this one, but it is higher voted and it has the higher voted answers. Commented Dec 10, 2020 at 8:47

Hints:

1) There is only one possible abelian group of order $\,30\,$

2) Any group $\,G\,$ of order 30 has a subgroup $\,H\,$ of order $\,15\,$, which is normal and abelian -- in fact, cyclic -- (why and why?), and thus $\,G\cong H\rtimes Q\,$ , for some subgroup $\,Q:=\langle\,q\,\rangle\,$ of order two.

Since $\,\operatorname{Aut}(H)\cong C_2\times C_4\,$ (why?) , there are at least four possible homomorphisms $\,Q\to\operatorname{Aut}(H)\,$ , all of them convolutions: (i) mapping $\,q\,$ to the generator of the factor $\,C_2\,$ , (ii) to $\,p^2\;,\;\;p=$ the generator of $\,C_4\,$ , and (iii) to the element $\,(q,p)\in C_2\times C_4\,$ (the trivial homomorphism gives the abelian group we already had before).

Check the above three non-trivial homomorphisms give three non-isomorphic groups of order $\;30\;$ .

• Does $Q$ is normal?
– Andy
Commented Feb 26, 2013 at 4:58
• Not in general, @AnindyaGhatak . In fact, it is normal iff $\,G\,$ is a direct product, and since both $\,H,Q\,$ are abelian we'd be back in case (1). Commented Feb 26, 2013 at 5:01
• in (ii), mapping q to p won't give a homomorphism. Commented Nov 16, 2013 at 15:12
• @DonAntonio: I'm sorry. I didn't realize that you wrote "Hints:" in the first line. Commented Nov 16, 2013 at 19:05
• @RohanRajagopal Whenever we have a semidirect product as in my answer, there exisats a homomorphism from the right factor into the automorphism group of the left factor. This is the explanation why there's stuff about automorphism groups in my answer. Commented Nov 26, 2018 at 13:18

There is a general description of groups with cyclic Sylow subgroups:

Marshall Hall, The Theory of Groups - Theorem 9.4.3.

• Here is a reference to this result at wikipedia.
– Ben
Commented Mar 11, 2018 at 2:09