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I was reading this page about the following result :

If a group has a faithful irreducible representation over any field of characteristic zero, then the center of the group is cyclic.

The said page has proof, but I am looking for a published book/paper, if available, for referring to.

Any help will be greatly appreciated.

Thanks in advance.

usermath
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1 Answers1

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This basically cycles back to the fact that a finite subgroup of the multiplicative group of a field of characteristic zero is cyclic. See I.M. Isaacs, Character Theory of Finite Groups, Lemma (2.27). And also here.

Let me explain it in more detail. First, if $\mathfrak{X}$ is a representation affording the character $\chi$, then $ker(\mathfrak{X})=ker(\chi)$. Here $ker(\mathfrak{X})=\{x \in G: \mathfrak{X}(x)=I\}$ and $ker(\chi)=\{x \in G: \chi(x)=\chi(1)\}$, see also Lemma (2.19) of Isaacs' book.

Now if $\chi$ is faithful then $ker(\chi)=\{1\}$. Hence Lemma (2.27) (d) and (f) will give you that $Z(G)$ must be cyclic.

Nicky Hekster
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