2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ To start with, did you find $Z(G)$? What is it? $\endgroup$ – Bungo Dec 27 '16 at 16:47
  • $\begingroup$ It is $\{1, a^4\}\times C_2$ $\endgroup$ – Maria Dec 27 '16 at 16:50
  • $\begingroup$ What is the order of the quotient group? What does that tell you about the quotient group? $\endgroup$ – Steve D Dec 27 '16 at 16:51
  • 1
    $\begingroup$ 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 ${\mathtt {UpperCentralSeries}}$ $\endgroup$ – Derek Holt Dec 27 '16 at 17:08
  • 1
    $\begingroup$ 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. $\endgroup$ – Derek Holt Dec 27 '16 at 17:42
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Dec 27 '16 at 17:56
  • 1
    $\begingroup$ And, sorry, I have grown up believing that $D_n$ is the group of symmetries of a regular $n$-gon :-) $\endgroup$ – Jyrki Lahtonen Dec 27 '16 at 17:57
  • $\begingroup$ @JyrkiLahtonen: How do you get cyclic of order 2? $a^2$ has order $4$. I think maybe you are thinking of $Z_2/Z_1$? $\endgroup$ – Steve D Dec 27 '16 at 17:59
  • $\begingroup$ @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 $Z(G/Z(G))$. $\endgroup$ – Bungo Dec 27 '16 at 18:07
  • $\begingroup$ 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) :-) $\endgroup$ – Jyrki Lahtonen Dec 27 '16 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.