# Thread: Nonabelian group of order p^3

1. ## Nonabelian group of order p^3

[Hungerford, p99 Q7]

Show that

If G is a nonabelian group of order $p^{3}$ (p prime), then the center of G is the subgroup generated by all elements of the form $aba^{-1}b^{-1} (a,b \in G)$.

2. Originally Posted by aliceinwonderland
If G is a nonabelian group of order $p^{3}$ (p prime), then the center of G is the subgroup generated by all elements of the form $aba^{-1}b^{-1} (a,b \in G)$.
Let $C$ be the subgroup generated by elements of the form $aba^{-1}b^{-1}$ i.e. commutator subgroup. By Burnside's lemma $Z(G)$, the center, is non-trivial. Therefore, $|Z(G)|\geq p$. But then the group $|G/Z(G)|$ has order $p\text{ or }p^2$ and therefore it is abelian. Thus, we must have that $C\subseteq Z(G)$. We are told that $G$ is not abelian therefore, $C\not = \{ 1 \}$ and $Z(G) \not = G$. Suppose that $|Z(G)|=p^2$ and choose $a\in G - Z(G)$. Then $C(a)$, the centralizer, contains $Z(G)$ properly, and is a subgroup of $G$. Thus, Lagrange's theorem forces $|C(a)| = p^3$. But that would mean the index of $C(a)$ in $G$ is $1$. This means that the conjugacy class of $a$ is just $\{ a \}$ and so $a\in Z(G)$ - a contradiction! Thus, we see that $|Z(G)| \not = p^2$. The only remaining possibility is that $|Z(G)| = p$. However, $C$ is a non-trivial subgroup of $Z(G)$ which forces $C = Z(G)$.