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These two exercises confuses me:

Let $G$ be a group of order $p^3$, where $p$ is a prime

  1. Show that if $|Z(G)| = p^2$ then $G/Z(G)$ is cyclic

  2. Show that if $G/Z(G)$ is cyclic, then $G$ is abelian

What? Isn't a group $G$ abelian iff $|Z(G)| = |G|$, which is obviously not the case when $|Z(G)| = p^2$?

José Carlos Santos
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mikkelmk
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    Yes you are right, so this exercise proves that $|Z(G)| \ne p^2$. – Derek Holt Jan 13 '18 at 22:31
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    What this exercise implies is that it is never the case that $|Z(G)| = p^2$ when $|G|=p^3$, which is a pretty sweet result, right ? – Maxime Ramzi Jan 13 '18 at 22:32
  • See also https://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point – lhf Jan 13 '18 at 22:34
  • In general, what you prove here is that the quotient of any group by its center can't never be non-trivial cyclic. – DonAntonio Jan 13 '18 at 22:40
  • It may also help with the slightly stronger formulation: If $H\subseteq Z(G)$ is a subgroup such that $G/H$ is cyclic, then $G$ is abelian. This makes it more clear that the situation can actually arise. – Tobias Kildetoft Jan 14 '18 at 08:27

2 Answers2

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Sure. And therefore it follows from 1. and 2. that, if $|G|=p^3$ for some prime $p$, then $\bigl|Z(G)\bigr|\neq p^2$.

José Carlos Santos
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  1. If $Z(G) = {p}^{2}: $ then $|G/Z(G)| = p$, then its only subgroups are {$e$} and $H$ with $|H|=p, \ so \ H = G$. Any non-identity element of $G$ has order $p$, then it generates all $G$, then it's cyclic.

  2. If $G/Z(G)$ cyclic, it means that $ \exists \ x \ \epsilon \ G : \ <xZ(G)>=G/Z(G) $

With this in mind we know that if ${g}_{1} , {g}_{2} \ \epsilon \ G$ then ${g}_{1}Z(G) = {(xZ(G))}^{a} = {x}^{a}Z(G)$. Then there exists ${z}_{1} \epsilon Z(G)$ such that ${g}_{1} = {x}^{a}{z}_{1}$. In the same way we can assume ${g}_{2} = {x}^{b}{z}_{2}$

${g}_{1}{g}_{2}= \ {x}^{a}{z}_{1} {x}^{b}{z}_{2}=... \\$ and since ${z}_{i} \epsilon Z(G)$ $ ...= \ {x}^{a}{x}^{b}{z}_{1}{z}_{2} = {x}^{b}{x}^{a}{z}_{2}{z}_{1}={x}^{b}{z}^{2}{x}^{a}{z}^{1}={g}_{2}{g}_{1}$

this proves that $G$ is abelian.

Melissa Robles Carmona
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