# Does the knowledge of $Z(G)$ and $G/Z(G)$ give the full group? [closed]

I'm wondering if by knowing the center $Z(G)$ as well as $G/Z(G)$ one can deduce G.

I thought that you should be able to write $G=Z(G) \times G/Z(G)$, because every element either lies in the center or it does not, and central elements can always be "separated" from the rest by commuting e.g. to the left. Yet a bit of experimenting with GAP has shown that this is clearly not true, however I don't see my mistake and would be very grateful if anyone could point it out to me.

• Look up "central extension" on wikipedia: en.wikipedia.org/wiki/Group_extension Basically, the 2nd cohomology group tells you how many distinct central extensions there are. Sometimes there's many, sometimes there's only one. – Ryan Budney Mar 13 '12 at 20:44
• This question currently has four votes to close, so I would like to register that I think this question is perfectly appropriate, and vote NOT to close. – Theo Johnson-Freyd Mar 14 '12 at 3:06
• There is a coffee-table book called something like "groups of order 64". It does begin the classification in terms of the centers and quotients and so on, so you're not too far off, but as you've seen in examples below, there's also the data of the extension, and not just the two component pieces. – Theo Johnson-Freyd Mar 14 '12 at 3:09
• Why has this question been closed? – Sean Eberhard Mar 14 '12 at 8:10
• |It is definitely not a research level question. It is undergraduate level, and entirely suitable for math s.e. – Derek Holt Mar 14 '12 at 8:25

No, the knowledge of $Z(G)$ and $G/Z(G)$ usually doesn't give $G$. As an example, the quaternion group $Q_8$ and the dihedral group $D_8$ have both center $\mathbb{Z}/2$ and quotient $\mathbb{Z}/2 \times \mathbb{Z}/2$.
The problem with your argument is that if $z$ is central, it may be of the form $z=g^n$ with a non-central $g$. Such a $g$ can't be factored into a central part and a non-central part.
No, one cannot. For instance the dihedral group $D_8$ has centre $Z(D_8)\cong C_2$ and $D_8/Z(D_8)\cong C_2\times C_2$, but $D_8\ncong C_2\times C_2\times C_2$.
I might paraphrase your argument as follows: Every $g\in G$ lies in exactly one coset $Z(G)h$ and therefore can be written uniquely as $g=zh$ where $z\in Z(G)$. When we multiply two elements $g=zh$ and $g^\prime = z^\prime h^\prime$ we obtain $gg^\prime = zz^\prime hh^\prime$. Since this corresponds to the group law in $Z(G)\times G/Z(G)$, we must have $G\cong Z(G)\times G/Z(G)$. The trouble is that the expression $g=zh$ is not at all unique: for instance $g=zz^\prime\cdot {z^\prime}^{-1}h$ for any $z^\prime\in Z(G)$ does just as well, and consistently choosing representations $g=zh$ for every $g\in G$ would be nothing but giving an isomorphism $G\cong Z(G)\times G/Z(G)$.
• That's true, good points. His intention seems to ask whether an isomorphism $G\cong Z(G)\times G/Z(G)$ always holds. (On the other hand, I guess clearly doesn't hold, as it would imply that $Z(G) \cong Z(G) \times Z(G/Z(G))$ always holds. – Sean Eberhard Mar 14 '12 at 8:08