2
$\begingroup$

Let $G$ be a group, and $H\leq G$ be a subgroup. When is $C_G (H)=Z(G)$?

Similar to this question, which is about the centralizer of an element rather than of a subgroup: When is the centralizer and the center are equal

$\endgroup$
2
  • $\begingroup$ How did you conclude that question is about the centralizer of an element? The OP never specified. $\endgroup$ Jan 7 '16 at 0:37
  • $\begingroup$ The answer was answered this way in the comments and the OP accepted it and said that it answered their question. $\endgroup$ Jan 7 '16 at 4:38
1
$\begingroup$

I do not think there is a general condition. What you can say is this.

(a) $Z(G)=C_G(H)$ if and only if $Z(H)=H \cap Z(G)$.

(b) Note that $Z(G)=\bigcap_{g \in G}C_G(g)$. And $g \notin Z(G)$ if and only if $Z(G) \subsetneq C_G(g) \subsetneq G$. Otherwise put, if $C_G(g)=Z(G)$, then, since $g \in C_G(g)$, we get $g \in Z(G)$, whence $C_G(g)=G$, so $G=Z(G)$, which means $G$ is abelian. And if $G$ is abelian, then of course $C_G(g)=G=Z(G)$.

$\endgroup$
3
  • $\begingroup$ Can you give me an example of a non-abelian group $G$ and subgroup $H$ satisfying condition (a)? $\endgroup$ Jan 7 '16 at 23:37
  • $\begingroup$ Yes sure. Take $G=S_3 \times C_2$, and $H=S_3 \times \{1\}$. $Z(G)=\{(1)\} \times C_2$, and $Z(H)=\{(1)\} \times \{1\}$. $\endgroup$ Jan 8 '16 at 7:25
  • $\begingroup$ @Joshua Meyers, if you deem my answers as the right ones, please tick it as such, complying with the habits on this site and having not too much question left unanswered. Thanks! $\endgroup$ Jan 11 '16 at 9:14
0
$\begingroup$

Here is another criterion, if you know Character Theory of Finite Groups. If not, consult the book, bearing the same title of Marty Isaacs, see for example here.

Proposition Let $H$ be a subgroup of the finite group $G$ $\chi \in Irr(G)$ and irreducible complex character and assume that the restriction $\chi_H \in Irr(H)$. Then $C_G(H) \subseteq Z(\chi)$.

Corollary If $H \leq G$, $\chi \in Irr(G)$ a faithful irreducible character, such that $\chi_H \in Irr(H)$ then $Z(G)=C_G(H)$.

So let us prove this. Let $\mathfrak{X}$ be an $\mathbb{C}$-representation that affords $\chi$. Let $x \in C_G(H)$. Since the restriction to $H$ is irreducible and $\mathfrak{X}(x)$ commutes with every $\mathfrak{X}(h), h \in H$, and it follows from Lemma (2.25) of the book (which is basically Schur's Lemma) that $\mathfrak{X}(x)=\varepsilon I$ for some root of unity $\varepsilon \in \mathbb{C}$. By taking traces it follows that $|\chi(x)|=\chi(1)$, that is $x \in Z(\chi)$.

Now assume that $\chi$ is faithful (meaning $ker(\chi)=1$). Since in general, $Z(G/ker(\chi))=Z(\chi)/ker(\chi)$, we obtain $Z(\chi)=Z(G)$. The proposition then yields $C_G(H) \subseteq Z(G)$. But of course the other containment, $Z(G) \subseteq C_G(H)$ is always true. Hence $Z(G)=C_G(H)$ as wanted. (Note that $Z(G)$ must be cyclic in this case since $Z(\chi)/ker(\chi)$ is always cyclic).

$\endgroup$

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.