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?
Best Answer
First, there are several errors in your calculations of the upper central series.
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...