# How do I find $Z_2(D_8 \times C_2)$?

I want to find the second center of $G=D_8 \times C_2$, where $D_8$ is the diheral group of order 16.

I'm having a hard time seeing how you would go about finding the preimage of $Z(G/Z(G))$ under the canonical epimorphism $\pi:G \to G/Z(G)$.

I'm not sure if there is a better method that one can use here.

Also, side question, is there a method to find this using GAP?

• To start with, did you find $Z(G)$? What is it? – Bungo Dec 27 '16 at 16:47
• It is $\{1, a^4\}\times C_2$ – Maria Dec 27 '16 at 16:50
• What is the order of the quotient group? What does that tell you about the quotient group? – Steve D Dec 27 '16 at 16:51
• Yes you can do it in GAP, but the group that you call $D_8$ is called ${\mathtt{DihedralGroup}}(16)$ in GAP. Use the GAP functions ${\mathtt {DirectProduct}}$ and ${\mathtt {UpperCentralSeries}}$ – Derek Holt Dec 27 '16 at 17:08
• The function $\mathtt{Projection}$ can be used to define the projection of a direct product onto its direct factors. But I had to look that up in the GAP manual, and you could just as easily do that yourself. – Derek Holt Dec 27 '16 at 17:42

Let's use the regular notation, and call your group $D_{16}\times C_2$.

Now $D_{16}$ is the dihedral group of order $16$, containing a cyclic subgroup of order $8$ (let's say generated by $a$) and an involution that inverts $a$ (let's call it $b$).

The center of $D_{16}$ is just the power of $a$ that -- when inverted -- is equal to itself. Thus the center of $D_{16}$ is generated by $a^4$.

The quotient $D_{16}/Z(D_{16})$ is still generated by $a$ and $b$, but the image of $a$ (let's call it $\bar{a}$) has order $4$ now. That is, this quotient is $D_8$.

The center of this quotient is generated by $\bar{a}^2$, because mod $4$, $2$ and $-2$ are the same. Back up in $D_{16}$, this means $Z_2$ is generated by $a^2$.

Thus $Z_2(D_{16}\times C_2) = \langle a^2\rangle\times C_2$.

• If $Z(D_8\times C_2)=\langle a^4\rangle \times C_2$, then $G/Z(G)\simeq D_8/\langle a^4\rangle\simeq D_4$. And $Z(D_4)=\langle a^2\rangle$, no? So the answer should be cyclic of order two? – Jyrki Lahtonen Dec 27 '16 at 17:56
• And, sorry, I have grown up believing that $D_n$ is the group of symmetries of a regular $n$-gon :-) – Jyrki Lahtonen Dec 27 '16 at 17:57
• @JyrkiLahtonen: How do you get cyclic of order 2? $a^2$ has order $4$. I think maybe you are thinking of $Z_2/Z_1$? – Steve D Dec 27 '16 at 17:59
• @JyrkiLahtonen The center of $D_4$ is cyclic of order $2$ (I wouldn't write $\langle a^2\rangle$ for this as it reuses $a$ to mean something new). So that gives us the center of the quotient group $G/Z(G)$. But this isn't the second center of $G$. To get that, we have to find the corresponding subgroup of $G$ whose image under the natural homomorphism is $Z(G/Z(G))$. – Bungo Dec 27 '16 at 18:07
• Ok. I thought you were looking for the group $Z(G/Z(G))$ rather than its preimage in $G$. Should have asked (or read more carefully) :-) – Jyrki Lahtonen Dec 27 '16 at 19:00