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This is probably stupid question, but I can't see the difference between the two subgroups:

$$C_G(A)=\{g\in G| gag^{-1}=a,\forall a\in A\}$$ $$Z(G)=\{g\in G| ga=ag,\forall a\in G\}$$

Is the difference that the centralizer takes a subset $A\in G$ and the center always uses the entire group $G$?

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  • $\begingroup$ Yes, $C_G(G) = Z(G)$. $\endgroup$ Jun 22, 2015 at 2:47
  • $\begingroup$ @Stefan Thanks and where does the normalizer stand? $\endgroup$ Jun 22, 2015 at 2:49
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    $\begingroup$ Yes the centralizer generalizes the center because it's relevant for subgroups, not just the whole group. $\endgroup$ Jun 22, 2015 at 2:49
  • $\begingroup$ As for the normalizer $Z(G) \subseteq C_G(A) \subseteq N_G(A)$ for every subgroup $A \le G$ and $N_G(A)$ is the maximal subgroup of $G$ in which $A$ is normal. $\endgroup$ Jun 22, 2015 at 2:52
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    $\begingroup$ I'm not sure whether this is standard notation, but in all my group theory courses $A \le G$ meant that $A$ is a subgroup of $G$, while $A \unlhd G$ meant that $A$ is a normal subgroup of $G$. $\endgroup$ Jun 22, 2015 at 4:50

2 Answers 2

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Converting the comments into an actual answer:

Yes, we have the special case $C_G(G)=Z(G)$, so the notion of centralizer can be thought of as a generalization of centers, since the centralizer $C_G(A)$ works for every subset $A$ of $G$ (note that it doesn't have to be a subgroup).

As for the normalizer, $Z(G)\subseteq C_G(A)\subseteq N_G(A)$ for every subgroup $A\leq G$, and $N_G(A)$ is the maximal subgroup of $G$ in which $A$ is normal. The normalizer can also be defined for an arbitrary subset, and the above still holds other than that we must now say that $\langle A \rangle$, the subgroup generated by $A$, is normal in $N_G(A)$, and the normalizer is maximal with respect to this property.

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  • $\begingroup$ Here,can we say $C_G(A)$ to be the centre of A? $\endgroup$
    – Styles
    Jun 26, 2016 at 22:34
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    $\begingroup$ @PKStyles No. Here $C_G(A) = \{ g \in G \ : \ ga=ag \ \forall a\in A \}$. It's the pointwise stabilizer of $A$ in $G$. It contains the center of $A$ when $A$ is a (sub)group, but can be strictly larger. Consider the case when $A$ is a subgroup (or even just subset) of $Z(G)$, for example. Then $C_G(A)=G$. $\endgroup$ Jun 27, 2016 at 6:40
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The normalizer of a subset of a group is the part of the group that commutes with the subset (but not necessarily holding the element of the subset fixed): $g$ in the normalizer of $H \subset G$ means $gH = Hg$, or more specifically, $gh_1 = h_2 g$ for some $h_1, h_2 \in H$.

The centralizer of a subset of a group is the part of the group that commutes with the subset elementwise (i.e., holding the element of the subset fixed): $g$ in the centralizer of $H \subset G$ means $gh = hg$ for all $h \in H$. Clearly, commuting elementwise implies commuting with the subset, so the normalizer of a subset contains the centralizer of that subset.

The center of a group is the part of the group that commutes with everything in the group. Commuting with everything implies commuting with elements of some subset, so the centralizer of a subset contains the center of the group.

Putting these together: $Z(G) \subset C_G(H) \subset N_G(H)$.

Regarding "$\leq$"... The centralizer and normalizer are defined on subsets. It may turn out that a subset is also a subgroup, but this is not required. (It will essentially always be the case that the subsets are subgroups once you are past the material introducing these definitions.) This is why I have used "$\subset$" in the above.

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