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 $125.$

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    $\begingroup$ By the usual generalized statement of the Sylow theorems we also know that there is a subgroup of order $25$. $\endgroup$ Commented Dec 15, 2019 at 18:54
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    $\begingroup$ It's a textbook result that there are two non-Abelian groups of order $p^3$ up to isomorphism for each prime $p$. $\endgroup$ Commented Dec 15, 2019 at 18:56
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    $\begingroup$ 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 $\endgroup$ Commented Dec 15, 2019 at 18:59
  • $\begingroup$ Thank you, how can I know there exists subgroup of order 25? $\endgroup$
    – able20
    Commented Dec 15, 2019 at 19:02
  • $\begingroup$ 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 $5$. $\endgroup$ Commented Dec 15, 2019 at 19:07

1 Answer 1


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.


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