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For a while I thought that, 'intuitively,' $G/Z(G)$ where $G$ is nonabelian with nontrivial center would have trivial center - that is, it would be 'totally nonabelian.' I realize now that this is not necessarily true, as in cases where $G/Z(G)$ has order $p$ or $p^2$ where $p$ is prime.

I also realize that what fed this idea and made it feel intuitive was the use of phrases about "modding out" or "quotienting out" normal subgroups. To me, this phrasing makes it sound like we're sort of 'removing' that subgroup. And indeed we are, in some cases. $\mathbb{Z}_6$ has elements of orders 1, 2, 3, and 6. Taking $\mathbb{Z}_6 / \{0,3\}$ (that is, 'quotienting out' a subgroup of order 2) we obtain $\mathbb{Z}_3$, and have effectively 'removed' the elements of order 2, since $\mathbb{Z}_3$ has only elements of orders 1 and 3.

So my question is: are there specific conditions under which $G/Z(G)$ (as described above) has trivial center?

More generally, can we say anything about when taking the quotient by a normal subgroup actually 'removes' the elements that characterize that subgroup?

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Well, for one, if your group is the direct product of an abelian and a nonabelian group with trivial center, then $G/Z(G)$ as trivial center. For an example where taking the quotient actually removes the elements characterizing the subgroup, note that the quotient by the commutator subgroup always gives an abelian group. –  Carl Feb 23 '12 at 21:11
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I don't understand your comment about $p$ and $p^2$: $G/Z(G)$ never has prime order. –  Chris Eagle Feb 23 '12 at 21:24
    
Ah yes, I didn't think of that. If $G/Z(G)$ has prime order then it is cyclic and so $G$ is abelian, but then $G=Z(G)$ and $G/Z(G)$ is trivial, contradicting $|G/Z(G)|=p$. –  Alex Petzke Feb 24 '12 at 2:53

2 Answers 2

up vote 8 down vote accepted

The pullback of the center of $G/Z(G)$ is called the "second center of $G$", $Z_2(G)$; it is the second term of the upper central series of $G$, which in turn is intimately connected with the lower central series of $G$ and to the notion of nilpotency.

Thus, $G/Z(G)$ has trivial center if and only if $Z_2(G)=Z(G)$. For a nilpotent group, this happens if and only if $G$ is abelian, in which case $Z(G)=G$. In particular, if $G/Z(G)$ is of prime power order, and not trivial, then $Z_2(G)$ is strictly larger than $Z(G)$.

There are many conditions that guarantee this; for example, if $G=[G,G]$, then $G/Z(G)$ is centerless.

Now, I think you are viewing quotients wrong; we are not "removing" the subgroup, we are trivializing it: we are making everything in the subgroup not matter (rather than "not being there"). If you view $Z(G)$ as the collection of all elements that commute with everything, then $Z_2(G)$ is the collection of all elements that commute with everything up to elements that "don't matter" (namely, the elements of the center, which commute with everything). That is, $g\in Z_2(G)$ if and only if for every $x\in G$ there exists $z\in Z(G)$ such that $gx = xgz$. If you consider elements of $Z(G)$ to "be trivial", "not matter" (which is what you do when you take the quotient), then $g$ "essentially" commutes with $x$.

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Can you describe or reference any of these conditions? Other than $G=G'$. –  zibadawa timmy Dec 8 '13 at 3:08

Certain properties of groups are closed under extension. That is, if $N$ is a normal subgroup of $G$ and both $N$ and $G/N$ have that property, then so does $G$.

So for example, the properties of being a $p$-group for a specific prime $p$, or of being solvable, or of being a torsion group are closed under extension, whereas the properties abelian and nilpotent are not.

For a property that is closed under extension, if there is a largest normal subgroup $N$ of $G$ having that property (which will always be the case for finite groups), then $G/N$ will have only the trivial normal subgroup with that property.

So, for a finite group $G$ and prime $p$, there is a largest normal $p$-subgroup, the so-called $p$-core $O_p(G)$, and $G/O_p(G)$ has trivial $p$-core. Similarly for the largest normal solvable subgroup.

For infinite groups it is more complicated, because there might not be a maximal normal $p$-subgroup or solvable subgroup.

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