# Center of non-abelian group of order $17^4$

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.

• What does it mean to "find its center"? – Servaes Feb 4 at 19:03
• set isomorphic to center – user389231 Feb 4 at 19:05
• 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 – DonAntonio Feb 4 at 19:18
• There are $10$ different nonabelian groups of order $17^4$. Which one is $G$? – Robert Israel Feb 4 at 19:56
• This is not a valid problem because it does not have a unique solution. – Derek Holt Feb 5 at 8:56

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 $$n^{17} \equiv 1 (\text{mod } {17}^2)$$