0

I was trying to show that if $G$ has order $p^3$ and is not abelian, then $Z(G)$ is the commutator subgroup.

Working out one of the cases, we get that if

$$|Z(G)|=p^2 \implies |G/Z(G)|=p \implies G/Z(G) \text{ is abelian}$$

I am trying to prove that $G$ is abelian and get to a contradiction. I know the groups $G/Z(G)$ and $Z(G)$ are of a completely different nature but if it was possible to bring $G/Z(G)$ to a subgroup $K$ of $G$ that is isomorphic to $G/Z(G)$ by $\varphi: G/Z(G) \to K$ then we could write $G=\varphi(G/Z(G))\times Z(G)$ and since both factors are abelian, $G$ would be abelian.

I wonder if it makes any sense in this example but also in a more general scenario.

Gabriela Martins
  • 459
  • 2
    Why must $\mu(G/N)$ be a subgroup? Such a $\mu$ need not be a homomorphism. (Consider $G=\mathbb{Z}, N = 2\mathbb{Z}$.) – Randall May 14 '25 at 01:20
  • 2
    This is also a little ridiculous because $\mu \circ \pi = \mathrm{id}$ implies just in set theory that $\pi$ is injective. But if the quotient is injective then $N$ is trivial. – Randall May 14 '25 at 01:22
  • 1
    So I guess the answer to your question is "always," since your assumptions show that $N$ is trivial so $G/N \cong G$ and your desired isomorphism just says $G \cong G \times {e}$. – Randall May 14 '25 at 01:26
  • I forgot to write that $\mu$ was a group homomorphism, my bad – Gabriela Martins May 14 '25 at 01:43
  • 1
    It doesn't matter because $N$ is still forced to be trivial, making the question itself trivial. This has nothing (nearly) to do with $\mu$ being a homomorphism, but your imposed condition that $\mu \circ \pi = \mathrm{id}$. – Randall May 14 '25 at 01:44
  • I edited the question now to better reflect what I was thinking – Gabriela Martins May 14 '25 at 01:55
  • Are you trying to prove that any group of order $p^3$ is abelian? That's false. – jjagmath May 14 '25 at 01:58
  • 1
    Without more information, it is impossible to prove that $G$ is abelian. Why don't you directly put the exercise you are working on? – azif00 May 14 '25 at 02:03
  • I was trying to get that $G$ was abelian and therefore it would be a contradiction that $H$ has order $p^2$ because the question stated that $G$ was not abelian – Gabriela Martins May 14 '25 at 02:05
  • What is $G$? What is $H$? What are the assumptions? What do you want to prove at the end? – azif00 May 14 '25 at 02:07
  • I added the exercise to the question – Gabriela Martins May 14 '25 at 02:09
  • $|G/Z(G)|=p$ implies that $G/Z(G)$ is cyclic. Then check this out: https://math.stackexchange.com/questions/63087/if-g-zg-is-cyclic-then-g-is-abelian – azif00 May 14 '25 at 02:13
  • yes, I used this to finish the exercise but I am more interested if it is possible to construct that direct product (and if not, why) rather than the exercise itself – Gabriela Martins May 14 '25 at 02:17
  • @GabrielaMartins Let $K \lhd G$ and $H \leq G$. If any $g \in G$ can be uniquely expressed as $g = kh$ for $k\in K, h\in H$, $G$ is called the semidirect product of $K$ and $H$. Under such circumstance, one can show that $G/K \cong H$, and one can make $H$ a system of coset representatives. If we assume further that $K$ and $H$ commute with each other, than the semidirect product becomes direct product. Given our experiences on vector spaces, it is tempting to decompose a group as the semidirect product of two proper subgroups. However, this is not always possible. – zyy May 15 '25 at 03:11
  • @GabrielaMartins As I understand, you are trying to write $G$ as a semidirect product using the following procedure: choose a set $S$ of representatives of $G/K$, such that $S$ happens to be a subgroup, and $G = K \rtimes S$ is the semidirect product (in your case, $K = Z(G)$ commutes with any subset, so $G = K \times S$). I myself tried to do this when I was still a beginner in group theory, but later found out that such a procedure does not work for a large class of groups. Take $\mathbb{Z}_8 = \langle g \rangle$, and the subgroup $K$ generated by $g^2$ as an example... – zyy May 15 '25 at 03:23
  • @GabrielaMartins ...The only subgroup of order $2$ is $\langle g^4 \rangle$, and one easily verifies that $K · \langle g^4 \rangle = K < \mathbb{Z}_8$. So one fails to find a subgroup isomorphic to $\mathbb{Z}_8/\langle g^4 \rangle$. Fortunately, you may find out from my first comment that whether $G$ is abelian is somehow independent of whether it can be expressed as a semidirect product. Actually, in your case it suffices to show that one can choose the representatives of $G/Z(G)$ to be powers of a single element. – zyy May 15 '25 at 03:29
  • @GabrielaMartins By the way, one of the many things makes the structure of a finite group $G$ more complicated than, say, vector spaces, is exactly the fact that $G$ cannot always be expressed as a semidirect product. That means, given groups $K, H$, there could be more than one groups $G$'s, not isomorphic to each other, such that $K \lhd G$ and $G/K \cong H$ (in contrast, we can only construct one vector space $V$ up to isomorphism if we know its subspace $W$ and the structure of $V/W$)... – zyy May 15 '25 at 03:41
  • @GabrielaMartins ...The so-called extension problem studies all possible structure of $G$ given $K$ and $H$, and you will find out one of the many key ideas to solve this problem is to study the obstacle that prevents $G$ to become a semidirect product, namely the factor set (also called $2$-cocycle). – zyy May 15 '25 at 03:44
  • @GabrielaMartins There is a typo in my third comment. By saying 'So one fails to find a subgroup isomorphic to $\mathbb{Z}_8/\langle g^4\rangle$', I actually mean 'So one fails to find a subgroup which happens to be a set of representatives of $\mathbb{Z}_8/K$'. In fact, it can be shown that such a procedure is the only way to construct a semidirect product, so one may consider it an equivalent characterization of semidirect product. For references, see section 2-3 in chapter 10, Rotman's Advanced modern algebra, 1st edition. – zyy May 16 '25 at 07:39

1 Answers1

1

You can use the fact that the commutator subgroup is the smallest normal subgroup that makes the quotient abelian. That characterizes the commutator subgroup.

Nicky Hekster
  • 52,147
MrGran
  • 775