# How many groups of order $125$ up to isomorphism?

What is the basic idea?

Normally for classification, I use Sylow theorem to use semidirect product, but since $$125 = 5^3$$, I cannot use it.

All I know is that $$Z(G)$$ cannot be identity and from the cauchy theorem I necessarily have subgroup of order $$5$$, but I don't seem to use these facts to figure out the ways to classify group of order $$125.$$

• By the usual generalized statement of the Sylow theorems we also know that there is a subgroup of order $25$. Commented Dec 15, 2019 at 18:54
• It's a textbook result that there are two non-Abelian groups of order $p^3$ up to isomorphism for each prime $p$. Commented Dec 15, 2019 at 18:56
• The abelian case is easy. For the non-abelian case the result described by Lord Shark the Unknown is explained in one of Keith Conrad's blurbs Commented Dec 15, 2019 at 18:59
• Thank you, how can I know there exists subgroup of order 25? Commented Dec 15, 2019 at 19:02
• You can look at the order of elements: try to show that if the group is not abelian you cannot have all elements of order $1$ and $5$. Commented Dec 15, 2019 at 19:07

Let $$p$$ be a prime number. Then there are, up to isomorphism, five groups of order $$p^3$$. These include three abelian groups and two non-abelian groups. The nature of the two non-abelian groups is somewhat different for the case $$p = 2$$.
The three abelian groups correspond to the three partitions of $$3$$.
The two non-abelian groups: For the case $$p = 2$$, these are dihedral group:$$D_{8}$$ and quaternion group. For the case of odd $$p$$, these are unitriangular matrix group:$$UT(3,p)$$ and semidirect product of cyclic group of prime-square order and cyclic group of prime order.