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$p$ prime number, $n$ a non-negative integer, $G$ group.

(a) $\forall G$ with $|G| = p^{n}$ cyclic $\Leftrightarrow$ $n \leq 1$.

(b) $\forall G$ with $|G| = p^{n}$ abelian $\Leftrightarrow$ $n \leq 2$.

My try:

a) "$\Leftarrow$": $n \leq 1$, hence $G \cong \mathbb{Z}_{p}$ and hence $G$ is cyclic.

"$\Rightarrow$": Say $n = 2$. As $G$ is cyclic, $Z(G)$ is non-trivial and $\exists g\in Z(G)$, $g \not= e$.

  • $g$ generates the whole group $\Rightarrow$ $G \cong \mathbb{Z}_{p^{2}}$ and thus $G$ cyclic

  • $g$ generates a subgroup of order $p$. Then $\exists h \notin Orbit(g)$ and the Order of the Subgroup generated by $g$ and $h$ has order $p^{2}$. (Also $g$ and $h$ commute as $g \in Z(G)$ and hence $G$ is abelian.) Thus $G$ is not generated by a single element and $G \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p}$. So we can find for $n>1$ groups generated by more than one element. And if we want all all groups of Order $p^{n}$ to be cyclic, $n$ has to be 1.

b) "$\Leftarrow$": Case $n=1$ and $n=2$ hold because of a) and the fact that cyclic $\Rightarrow$ abelian.

"$\Rightarrow$": Be $n = 3$. $Z(G)$ is non-trivial. So $Z(G) = p$, $p^{2}$ or $p^{3}$. For $p^{3}$ $G$ is indeed abelian. For $p^{2}$ as well. Because $|Z(G)| = p^{2} \Rightarrow |G/Z(G)| = p \Rightarrow$ $G$ is abelian.

$|Z(G)| = p$ then $Z(G) \cong \mathbb{Z}_{p}$ and $|G/Z(G)|=p^{2}$, so that $G/Z(G) \cong \mathbb{Z}_{p^{2}}$ (then $G/Z(G)$ is cyclic and hence $G$ abelian), or $G/Z(G) \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p}$. Then $G/Z(G)$ is non-cyclic and $G$ non-abelian.

For $n>2$ we can always find a group $G$ with $G/Z(G)$ like this. Hence if we want all groups with order $p^{n}$ to be abelian, then $n$ needs to be $\leq 2$.

Is that OK that way? Best, Sara :)

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1 Answer 1

a) "$\Rightarrow$" is a bit convoluted. Why not simply check that $G=(\mathbb Z_p)^2$ is not cyclic because all elements have order $\le p$?

b) "$\Leftarrow$" You should elaborate more on why $|G|=p^2$ implies that $G$ is abelian.

"$\Rightarrow$" You don't really show that $G$ with $|G|=p^3$ and $|Z(G)|=p$ does actually exists - what you write is more like a wishlist for such $G$. Also, you seem to think that $|Z(G)|=p^2$ and $|G|=p^3$ implies $G$ is abelian. If so, it does only becaus eno such group exists for $G$ being abelian is equivalent to $G=Z(G)$.

For an explicit example however, $\mathbb Z_p$ operates on $\mathbb Z_{p^2}$ by multiplication with $1+px$, that is we obain a semidirect product - or you can directly define a group structure on the set $\mathbb Z\times \mathbb Z_{p^2}$ by $(x,y)\circ(u,v)=(xu,y(1+pu)v)$ (verify that this is indeed a group!) and observe that $(0,1)\circ(1,1)=(0,p+1)$ whereas $(1,1)\circ(0,1)=(0,1)$.

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Ah, with a) $\Rightarrow$, I didn't think of that... So it's enough to say for the $n$-case: in $(\mathbb{Z})^{n}$ all Elements have order $\leq p$ but for a cyclic group we have the property that all the elements have the order $= |G|$? To b) $\Rightarrow$, I think I don't understand the group yet.. For the neutral element I guess I can just take $(1,1)(0,1)$. But already for the closure I don't get How I multiply the elements $(x_{1}u_{1},y_{1}(1+pu_{1})v_{1})(x_{2}u_{2},y_{2}(1+pu_{2})v_{2})$. Is that $(x_{1}x_{2}u_{1}u_{2}, y_{1}y_{2}(1+pu_{1}u_{2})v_{1}v_{2})$ or –  Martha Oct 14 '12 at 21:11
    
more complicated: $(x_{1}x_{2}u_{1}u_{2}, y_{1}y_{2}(1+2pu_{1}u_{2}+p^{2}u_{1}u_{2})v_{1}v_{2})$? And how do I find that this is in the group? Thanks for the Help so far!! –  Martha Oct 14 '12 at 21:14

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