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So I was working through some problems in a book on $p$-groups and noticed that $p$-groups have some really nice properties. So I started computing what the values of $|G/Z(G)|$ for $p$-groups. I decided to see what it would be and found that it cannot attain the value of $p$. I am interested in the whethere $|G/Z(G)|$ can attain all other powers of $p$?

In general I am interesting in the values that $|G/Z(G)|$ can achieve.

Any help is greatly appreciated.

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    $\begingroup$ For example, any $n$ such that every group of order $n$ is cyclic (e.g. the $n$ such that $(n,\varphi(n))=1$) are not attainable. For then, $G/Z(G)$ would necessarily be cyclic, so that $G$ is abelian so that $|G/Z(G)|=1$ . I don't know if this is the only restriction though... This is if I am understanding your question correctly: characterize $n$ which occur as $|G/Z(G)|$ for some group $G$. $\endgroup$ Oct 5, 2013 at 0:10
  • $\begingroup$ PS (+1), interesting question! Also, the e.g. in my last question should be i.e. The numbers $n$ such that every group of order $n$ is cyclic are precisely the $n$ such that $(n,\varphi(n))=1$. $\endgroup$ Oct 5, 2013 at 0:11
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    $\begingroup$ The primes, and more generally products $p_1\ldots p_m$ such that $p_i\not\equiv 1\mod p_j$ for all $i,j$. Also, I'm a little confused by the indexing of your sets $\psi_n$ though. I see no dependence on $n$. $\endgroup$ Oct 5, 2013 at 0:19
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    $\begingroup$ Here are some comments, but not a complete solution. First, if $n=p_1^{\alpha_1}\cdots$ and all the $\alpha_i>1$, your results on $p$-groups imply $n\in\psi$. Second, if $n$ is even - say $n=2m$ - then the dihedral group $D_{2n}$ is such that $D_{2n}/Z(D_{2n})$ has order $n$. So all even numbers are in $\psi$. The odd $n$ case is much harder. $\endgroup$
    – user641
    Oct 7, 2013 at 7:37
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    $\begingroup$ There are certain numbers that cannot be in $\psi$: as already remarked, the cyclic numbers; also, abelian numbers with one or more of the $\alpha_i=1$ (as an example, $45\notin\psi$); there are other cases, but they seem harder to classify. Note that the much harder problem of finding groups that are central quotients of other groups is the classification of capable groups problem, of which there is much literature (a lot by our our own @ArturoMagidin). $\endgroup$
    – user641
    Oct 7, 2013 at 7:37

2 Answers 2

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The result you found that $|G/Z(G)|$ cannot be prime is essentially the only restriction for $p$-groups:

For every $k\geq 2$ and every $p\in\mathbb{P}$ there exists a finite $p$-group $G$ such that $|G/Z(G)|=p^k$.

Proof: Let $q:=p^{k-1}$ and consider the matrix group $\begin{pmatrix}1&\mathbb{F}_p&\mathbb{F}_q \\ &1&\mathbb{F}_q \\ && 1\end{pmatrix}$ which has order $p^{2k-1}$ and center of order $p^{k-1}$ (the top right corner) which gives us a center quoient of order $p^k$ as desired. QED.

Observe that by taking direct products of such groups one can easily write down examples of $|G/Z(G)|=n$ for every natural number $n>1$ in which every prime factor that occurs, occurs at least twice. This is because $Z(G_1\times G_2)=Z(G_1)\times Z(G_2)$.

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In connection of this question it might be interesting to note that bounding the size of $G/Z(G)$ for a given finite group $G$ has some interesting history. For a given finite group Jordan's theorem gives an upper bound on the size of $G/Z(G)$ in terms of the degree of the smallest faithful representation of $G$.

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