Groups with large centers

Question

Consider any finite $2-$group $G$ and let $|G:C(G)|=4$. What can be said about $G$? is there any classification of these groups?

By $C(G)$ I mean center of $G$ and by $|G:H| = |G|/|H|$.

Answer 1

Remark: $2$-groups with $G/Z(G)\cong C_2\times C_2$ may still be too hard to classify.

The second remark, by the Lord, can be found here:

If $G$ is a group and $[G:Z(G)]=4$, show that $G/Z(G)$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

Answer 2

This is a very interesting question, but @Dietrich Burde is probably correct that this is likely not an easy problem. Your condition implies that the group has class $2$, and I'm not aware of a general classification for $2$-groups of class $2$, though your condition is admittedly stronger.

For reasons I don't fully understand, there seems to have been better progress classifying groups by coclass. (If $G$ has order $p^n$ and class $c$, its coclass is $n-c$, which is the difference between its derived length and nilpotency class.) See here, for example. Depending upon where you are coming at this from, that might be an interesting avenue for you to look into.

Having said that, there is a complete classification of two-generator $p$-groups of class $2$, described here, which is fairly recent. You might be able to fish out those among them with central quotient $C_2\times C_2$.

Answer 3

This class of groups is quite restricted and should admit a full classification, or something very close to it.

Let $G$ be a group with center of index $4$, and $Z$ be its center. Then $G$ is generated by $Z$, together with two extra elements, say $x$ and $y$. Moreover, $G$ is completely determined by the choice of $Z$ together with the choice of three additional elements of $Z$, corresponding to $x^2$, $y^2$, and $[x,y]$.

EDIT: Note that this is a duplicate of https://mathoverflow.net/questions/177693/classification-of-2-groups-with-center-of-index-4

In a comment there, Derek Holt claims that the only such groups are direct or central product of $D_8$ or $Q_8$ with an abelian $2$-group. That doesn't seem right though. There are examples of order $16$ that are not of this type, for example the non-trivial semidirect product $C_4\rtimes C_4$.

I think what is true is that it suffices to classify the case when $G=\langle x,y\rangle$, and then the general result can be obtained by taking central products with examples that arise in this case.

EDIT 2: Let $z=[x,y]\in Z$. Then $x^y=xz$ so $x=x^{y^2}=(xz)^y=xz^2$, so $z^2=1$. This shows that the derived subgroup has order $2$.