4
$\begingroup$

If $G$ is a $2$-group with cyclic center having order of $Z(G)=2^n$ and index of $|G/Z(G)|=4$, then how many groups of a particular order are possible?

I know that there are two possibilities for such a group:

  1. $G = \langle x,y,s \mid x^2 = y^2 = 1\rangle$ with $x,y \in G$ and $s$ is generator of center,
  2. $G = \langle x,y,s \mid x^2 = y^2 = s\rangle$, where $Z(G) = \langle s \rangle$ and $|G/Z(G)|=4$.

Further I know some properties of these groups:

  1. $x^2 \in Z(G)$ for every $x \in G$,
  2. $xy = yx$ or $xy = eyx$ for $e \in Z(G)$.

How to prove that these are the only two possibilities?

$\endgroup$
  • $\begingroup$ I think a more essential question is whether the case 2 determines the group up to isomorphism? Working on it... $\endgroup$ – Jyrki Lahtonen Jan 9 '16 at 9:57
  • 1
    $\begingroup$ @JyrkiLahtonen I have added a brief explanation to my answer. $\endgroup$ – Derek Holt Jan 9 '16 at 10:38
2
$\begingroup$

If $x^2 = s^{2k}$ for some $k$ then $(xs^{-k})^2=1$ so by replacing $x$ by $xs^{-k}$, we may assume that $x^2=1$.

If $(xs^{-k})^2=s$ so by replacing $x$ by $xs^{-k}$, we may assume that $x^2=s$.

So we may assume that $x^2=1$ or $s$ and similarly $y^2=1$ or $s$, and also $z^2=1$ or $s$, where $z=1$ or $s$. So by replacing $x$ or $y$ by $z$ if necessary, we can get either $x^2=y^2=1$ or $x^2=y^2=s$, giving the two possibilities.

To show that the groups are uniquely defined up to isomorphism, we also have to show that the commutator $[x,y] = x^{-1}y^{-1}xy$ is determined. Since $[x,y] \in Z(G)$, we have $[x,y]^2 = [x^2,y] = 1$. So (since $G$ is nonabelian) we must have $[x,y]= s^{2^{n-1}}$ in both cases.

$\endgroup$
  • $\begingroup$ @ Derek how to show that whenever x^2=1 or s then y^2 is equal to x^2 $\endgroup$ – Mittal G Jan 9 '16 at 10:16
  • $\begingroup$ @ Derek,ok got it. thanks $\endgroup$ – Mittal G Jan 9 '16 at 10:21
0
$\begingroup$

We know $G/Z(G)$ is not cyclic (or else $G$ is abelian), so $G/Z(G)$ is the Klein Four-Group, in which every element is self inverse, hence $g^2\in Z(G)$ for every $g$. Property 2 follows from the fact that $Z(G)$ must contain the commutator subgroup of $G$, because $G/Z(G)$ is abelian.

$\endgroup$

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.