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Let $G$ be group of order $17^4$. I have to find its center $Z(G)$ and $G/Z(G)$.

$|G|=17^4$.

Since $Z(G)$ is subgroup of $G$, the order of center divides the order of the group: $|Z(G)|=17^a$, where $a\leq4$.

$1^{\circ}$ $a=4$ $\Rightarrow |Z(G)|=|G|$. Since center of the group is Abelian, and the group is non-abelian, we have contradiction.

$2^{\circ}$ $a=3$ $\Rightarrow |G/Z(G)|=17$. This means that $G/Z(g)$ is cyclic, and then $G$ is Abelian. Contradiction.

What should I do with the cases $a=2$ and $a=1$?

And also, is this the right way to do this problem? Thank you.

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  • $\begingroup$ What does it mean to "find its center"? $\endgroup$ – Servaes Feb 4 at 19:03
  • $\begingroup$ set isomorphic to center $\endgroup$ – user389231 Feb 4 at 19:05
  • $\begingroup$ You have done right so far, and I think that both cases $\;|Z(G)|=17,\,17^2\;$ can happen... and this can be somehow messy, I believe. You may want to read arxiv.org/pdf/1611.00461.pdf $\endgroup$ – DonAntonio Feb 4 at 19:18
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    $\begingroup$ There are $10$ different nonabelian groups of order $17^4$. Which one is $G$? $\endgroup$ – Robert Israel Feb 4 at 19:56
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    $\begingroup$ This is not a valid problem because it does not have a unique solution. $\endgroup$ – Derek Holt Feb 5 at 8:56
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Actually, for all groups of order $17$ or $17^2$ there exist the examples of groups of order $17^4$, whose center is isomorphic to them.

For $C_{17} \times C_{17}$ it is $C_{17} \times (C_{17^2} \rtimes C_{17})$

For $C_{17^2}$ it is $C_{17^3} \rtimes C_{17}$

For $C_{17}$ it is $\langle x, y, z | x^{17^2} = e, x^{17} = y^{17} = z^{17}, x^{-1}yx = y^n, x^{-1}zx = z^n, y^{-1}zy = z^n \rangle$ where $n^{17} \equiv 1 (\text{mod } {17}^2)$

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