[Hungerford, p99 Q7]
Show that
If G is a nonabelian group of order $\displaystyle p^{3}$ (p prime), then the center of G is the subgroup generated by all elements of the form $\displaystyle aba^{-1}b^{-1} (a,b \in G)$.
[Hungerford, p99 Q7]
Show that
If G is a nonabelian group of order $\displaystyle p^{3}$ (p prime), then the center of G is the subgroup generated by all elements of the form $\displaystyle aba^{-1}b^{-1} (a,b \in G)$.
Let $\displaystyle C$ be the subgroup generated by elements of the form $\displaystyle aba^{-1}b^{-1}$ i.e. commutator subgroup. By Burnside's lemma $\displaystyle Z(G)$, the center, is non-trivial. Therefore, $\displaystyle |Z(G)|\geq p$. But then the group $\displaystyle |G/Z(G)|$ has order $\displaystyle p\text{ or }p^2$ and therefore it is abelian. Thus, we must have that $\displaystyle C\subseteq Z(G)$. We are told that $\displaystyle G$ is not abelian therefore, $\displaystyle C\not = \{ 1 \}$ and $\displaystyle Z(G) \not = G$. Suppose that $\displaystyle |Z(G)|=p^2$ and choose $\displaystyle a\in G - Z(G)$. Then $\displaystyle C(a)$, the centralizer, contains $\displaystyle Z(G)$ properly, and is a subgroup of $\displaystyle G$. Thus, Lagrange's theorem forces $\displaystyle |C(a)| = p^3$. But that would mean the index of $\displaystyle C(a)$ in $\displaystyle G$ is $\displaystyle 1$. This means that the conjugacy class of $\displaystyle a$ is just $\displaystyle \{ a \}$ and so $\displaystyle a\in Z(G)$ - a contradiction! Thus, we see that $\displaystyle |Z(G)| \not = p^2$. The only remaining possibility is that $\displaystyle |Z(G)| = p$. However, $\displaystyle C$ is a non-trivial subgroup of $\displaystyle Z(G)$ which forces $\displaystyle C = Z(G)$.