# prove a finite nilpotent group has supersolvable series using upper center series only

Let $$G$$ be a finite nilpotent group. If G has a normal series $$\langle e\rangle=G_0\leq G_1\leq ... \leq G_n=G$$ such that $$G_i\lhd G$$ and $$G_{i+1}/G_i$$ is cyclic for all $$i$$, then G is called supersolvable.

I want to prove that a finite nilpotent group $$G$$ has a normal series above whose factors are cyclic.

As $$G$$ is a nilpotent group, there must be a upper center series,

$$\langle e\rangle=Z(G)\leq Z_1(G)\leq\cdots\leq Z_n(G)=G$$

and $$G_{i+1}/G_{i}=Z(G/Z_i(G))$$. And each $$Z_i(G)$$ should be finite as $$G$$ is finite. I guess there must be some logic which allow me to insert a cyclic series inbetween the terms $$G_{i+1}$$ and $$G_i$$. But I wasn't get any idea. There was some existing solution like here, which claim, "A finite nilpotent group is the direct product of its Sylow subgroups". We haven't taught that, and I want to skip that too. Like, I want to solve that purely based on nilpotent help.

Any help will be appreciated. TIA

You use $$n$$ for two different things; it would be better to reserve it for the length of the upper central series, and use $$m$$ for the first normal series, or some other index...

Proceed by induction on $$n$$, the length of the upper central series.

If $$n=1$$, so $$G=Z(G)$$ is abelian, we can decompose $$G$$ into a direct sum of cyclic groups, $$G = C_1\oplus C_2\oplus\cdots\oplus C_k$$ and letting $$G_0$$ be trivial and $$G_i = C_1\oplus\cdots \oplus C_i$$, $$i=1,\ldots,n$$, you get a normal series in which each term is normal in the group and the consecutive quotients are all cyclic.

Assume the result holds for nilpotent groups whose upper central series has length $$k$$, and let $$G$$ have upper central series of length $$k+1$$. Let $$H=G/Z(G)$$, which is nilpotent with upper central series of length $$k$$, since $$Z_i(H) = Z_i(G/Z(G)) = Z_{i+1}(G)$$.

Thus, we have a normal series for $$H$$, $$\{1\} = H_0\lt\cdots \lt H_r=H,$$ with $$H_i\triangleleft H$$ for all $$i$$, and $$H_{i+1}/H_i$$ cyclic for $$i=0,\ldots,r-1$$.

Let $$N_i$$ be the subgroup of $$G$$ that contains $$Z(G)$$ and corresponds to $$H_i$$ under the canonical projection $$G\to G/Z(G)=H$$. By the correspondence theorem, $$N_i\triangleleft G$$ (since $$H_i\triangleleft H$$), and by the Third Isomorphism Theorem, $$\frac{H_{i+1}}{H_i} = \frac{N_{i+1}/Z(G)}{N_i/Z(G)} \cong \frac{N_{i+1}}{N_i},$$ so $$N_{i+1}/N_i$$ is cyclic for all $$i$$.

Now, since $$Z(G)$$ is abelian, it is a direct sum of cyclic groups. In addition, every subgroup of $$Z(G)$$ is normal in $$G$$. Write $$Z(G) = C_1\oplus\cdots \oplus C_m$$ with $$C_i$$ cyclic, and let $$M_0 = \{1\}$$, $$M_1=C_1$$, $$M_2=C_1\oplus C_2,\ldots, M_m=Z(G)$$. Then we have a normal series $$1 = M_0 \lt M_1\lt\cdots \lt M_m=Z(G)=N_0 in which each $$M_i$$ and each $$N_j$$ is normal in $$G$$, $$N_{i+1}/N_i$$ is cyclic for $$i=0,\ldots,r-1$$, and $$M_{i+1}/M_i$$ is cyclic for $$i=0,\ldots,m-1$$.

This shows $$G$$ is supersolvable.

• Thanks(+1) for your response, @arturomagidin. But I didn't see why should $M_i$ need to be normal in $G$? Because I guess it need not to be true that all cyclic subgroups are normal in the group. Can you clear this doubt? Thanks again Commented Nov 18, 2023 at 7:13
• @N00BMaster I justified that: "In addition, every subgroup of $Z(G)$ is normal in $G$." The $M_i$ are subgroups of $Z(G)$. Commented Nov 18, 2023 at 14:25