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Timeline for p-group with large center

Current License: CC BY-SA 3.0

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Jul 5, 2016 at 12:25 answer added yakov timeline score: -4
Jul 5, 2016 at 12:19 comment added yakov In the my post above such $p$-groups are characterized so other atgument are superfluous. Indeed, let $|G:{\rm Z}(G)|\p^2$ and $S\le G$ minimal nonabelian. Set $H=S\text{Z}(G)$. As $|H:\text{Z}(G)|\ge p^2=|G:\text{Z}(G)|$, we get $H=G$. Conversely, if $G=S\text{Z}(G)$, where $S$ is minimal nonabelian, then $\text{Z}(S)\le\text{Z}(G)$ and we conclude that $|G:\text{Z}(G)|=|S:\text{Z}(S)|=p^2$.
Jul 5, 2016 at 12:12 comment added yakov In my post above such groups are characterized. So all other argumant are auperfluous. Indeed, let $|G:\text{Z}(G)|=p^2$ and let $S\le G$ be minimal nonabelian. Set $H=S\text{Z}(G)$. As $H$ is nonabelian, $|H:\text{Z}(G)|\ge
Jun 25, 2016 at 16:11 comment added yakov A $p$-group $G$ satisfies the condition iff $G=S{\rm Z}(G)$, where $S\le G$ is minimal nonabelian.
Feb 25, 2013 at 4:39 answer added Arturo Magidin timeline score: 5
Feb 21, 2013 at 20:30 comment added Hamid Shahverdi Yes, You are right.
Feb 21, 2013 at 19:59 comment added Arturo Magidin @Hamid: No, there's many such groups, obtained by varying $A$. E.g., with $n=5$, you can have $A=C_{p}^3$, yielding a group of exponent $p$, or $A=C_{p^3}$, yielding a group with an element of order $p^3$.
Feb 21, 2013 at 19:39 answer added Ralph timeline score: 13
Feb 21, 2013 at 19:37 comment added Hamid Shahverdi Steve, For any integer $n\geq 4$ there exist only one group sa you make.
Feb 21, 2013 at 19:16 comment added Steve D If $H$ is the p-group of exponent $p$ and order $p^3$, and $A$ is any abelian group, then any central product of $A$ and $H$ has center $A$ of index $p^2$.
Feb 21, 2013 at 19:14 history edited Alexander Chervov
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Feb 21, 2013 at 17:22 history asked Hamid Shahverdi CC BY-SA 3.0