# $D_4 \times \mathbb{Z}_2$ different upper and lower central series

I wanted to find a group $$G$$ which had different upper and lower central series. Moreover, both different to one of his central series.

I have found that $$D_4\times \mathbb{Z}_2$$ is one of the counterexamples with the less nilpotency class. That is why I was trying to prove the fact above mentioned.

I have started computing the upper central series as follows:

$$Z_0(D_4\times \mathbb{Z}_2)=1$$

then I have defined $$Z_1(D_4\times \mathbb{Z}_2)$$ by the relation

$$\frac{Z_1(G)}{Z_0(G)} = Z(\frac{G}{Z_0(G)})$$

$$Z_1(G)= Z(D_4\times \mathbb{Z}_2) = Z(D_4)\times Z(\mathbb{Z}_2) =\{1,a^2\}\times \{1\}\cong \mathbb{Z}_2$$

I have continue and I have obtained

$$\frac{Z_2(G)}{\mathbb{Z}_2} = Z(\frac{D_4\times \mathbb{Z}_2}{\mathbb{Z}_2}) = Z(D_4) \cong \mathbb{Z}_2$$

Hence $$Z_2(D_4 \times \mathbb{Z}_2)= D_4 \times \mathbb{Z}_2$$, and we have

$$1 \leq \mathbb{Z}_2 \leq D_4 \times \mathbb{Z}_2$$

But my problem comes when I have to compute the lower central series and a central series.

I define $$\gamma_1(D_4 \times \mathbb{Z}_2)=D_4 \times \mathbb{Z}_2$$. Then I want to compute $$\gamma_2(D_4 \times \mathbb{Z}_2)=\gamma_2(D_4) \times \gamma_2(\mathbb{Z}_2)$$. But how can I compute the lower objects of $$D_4$$? Any help?

• Your calculation of $Z_2$ is incorrect, because the "$\mathbb{Z}_2$" that you have for $Z_1$ is not the second factor in the product $D_4\times\mathbb{Z}_2$; that copy of $\mathbb{Z}_2$ is embedded as a subgroup of $D_4$, so that $D_4\times\mathbb{Z}_2/Z_1(D_4\times \mathbb{Z}_2)$ is not $(D_4\times\mathbb{Z}_2)/(1\times\mathbb{Z}_2)$ (which would yield $D_4$ as you claim), but rather it is $(D_4\times\mathbb{Z}_2)/(\{1,a^2\}\times\{1\})) \cong (\mathbb{Z}_2\times\mathbb{Z}_2)\times\mathbb{Z}_2$. Commented Jan 25, 2019 at 18:01

You are correct that if $$A$$ and $$B$$ are groups, then $$Z(A\times B)=Z(A)\times Z(B)$$. So $$Z(D_4\times \mathbb{Z}_2) = Z(D_4)\times\mathbb{Z}_2$$. You are also correct that if we have $$D_4 = \langle a,s\mid a^4=s^2=1, sa=a^3s\rangle$$, then $$Z(D_4 = \{1,a^2\}$$. However, $$Z(\mathbb{Z}_2)=\mathbb{Z}_2$$, not $$\{1\}$$, so $$Z(D_4\times \mathbb{Z}_2) = \{1,a^2\}\times\mathbb{Z}_2$$. Thus, the quotient is isomorphic to $$(D_4/\{1,a^2\})\times(\mathbb{Z}_2/\mathbb{Z}_2)\cong (\mathbb{Z}_2\times\mathbb{Z}_2)\times\{1\}\cong \mathbb{Z}_2\times\mathbb{Z}_2$$. The center of this group is everything. Thus, the upper central series here is given by: $$\{1\}\times \{1\} \leq \{1,a^2\}\times \mathbb{Z}_2 \leq D_4\times \mathbb{Z}_2.$$ The middle term is isomorphic to $$\mathbb{Z}_2\times\mathbb{Z}_2$$, not to $$\mathbb{Z}_2$$.
Your next calculation, taking your computation of $$Z(D_4\times\mathbb{Z}_2)$$ as correct, doesn't work either. You state that the center is isomorphic to $$\mathbb{Z}_2$$; even assuming it were, it does not correspond to the subgroup $$1\times \mathbb{Z}_2$$, but rather it corresponded to a subgroup of $$D_4$$ times $$\{1\}$$. If that were the correct center, then the quotient would be isomorphic to $$(D_4/\{1,a^2\})\times (\mathbb{Z}_2/\{1\})$$, and not to $$(D_4\times \mathbb{Z}_2)/(1\times\mathbb{Z}_2)$$.
For the lower central series, you need to compute $$\gamma_1(D_4)$$. This is the subgroup generated by all elements of the form $$[x,y] = x^{-1}y^{-1}xy$$ with $$x,y\in D_4$$. If $$x$$ and $$y$$ commute, you get the identity. Note also that $$\gamma_1(D_4)$$ is the smallest normal subgroup such that $$D_4/\gamma_1(D_4)$$ is abelian. Since $$D_4$$ is nonabelian of order $$8$$, any normal subgroup of order $$2$$ will be the commutator (and hence the only normal subgroup of order $$2$$).
One commutator is $$[a,s]=a^{-1}s^{-1}as = a^3sas = a^3(sa)s = a^3a^3ss = a^2$$. So in fact, $$\gamma_1(D_4) = \{1,a^2\}$$. Thus, $$\gamma_1(D_4\times \mathbb{Z}_2) = \gamma_1(D_4)\times\gamma_1(\mathbb{Z}_2) = \{1,a^2\}\times\{1\}$$. Now notice that this is central, so $$\gamma_2(D_4\times \mathbb{Z}_2) = [\gamma_1(D_4\times\mathbb{Z}_2),D_4\times\mathbb{Z}_2) = \{1\}$$. Thus, the lower central series is given by $$\{1\}\times\{1\}\leq \{1,a^2\}\times\{1\} \leq D_4\times\mathbb{Z}_2.$$
This does give a group in which the lower and upper central series differ. However, you don't have "enough space" for a third central series between them, since there are no subgroups strictly between $$\{1,a^2\}\times\{1\}$$ and $$\{1,a^2\}\times \mathbb{Z}_2$$. If only you had a bit more "space" in that second factor...