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Why are the primitive characters on a $p$-group $G$ necessarily linear?

If $\chi$ is a primitive character with nontrivial kernel, then viewing it as a character on $G/\ker\chi$, it is primitive, so linear by induction as the base case is trivial to check.

So the difficulty is when $G$ has a faithful, primitive character. In this case, $Z(\chi)=Z(G)$, and $Z(G)$ is cyclic, and the unique largest abelian normal subgroup of $G$. (This is Corollaries 6.12, 6.13 in Isaacs' character theory text.)

By Clifford's Theorem, restricting $\chi$ to the center decomposes as $$ \chi_{Z(G)}=e\sum_{i=1}^t\theta_i $$ where the constituents $\theta_i$ form a $G$-orbit. But since any constituent $\theta$ is a character on $Z(G)$, then for any $g\in G$ and $x\in Z(G)$, $\theta^g(x)=\theta(g^{-1}xg)=\theta(x)$ since $x$ is central, so $\theta$ is fixed under the action of $G$, meaning $$ \chi_{Z(G)}=e\theta. $$ Then $$ \chi(1)=\chi_{Z(G)}(1)=e\theta(1)=e $$ since $\theta$ is linear. Is there a way to determine $e=1$?

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Hint: $p$-groups are monomial, that is, every irreducible character is induced by a linear character from some subgroup.

Let me give you a full proof that does not mention monomiality at all.

Theorem Let $G$ be a $p$-group and $\chi \in Irr(G)$ be primitive. Then $\chi$ is linear, that is $\chi(1)=1$.

We need two lemmata.

Lemma 1 Let $G$ be a $p$-group, then it contains a normal subgroup $A$ with $A=C_G(A)$ (note that such a subgroup must be abelian).
Proof Let $A$ be maximal among the abelian normal subgroups of $G$. Such an $A$ exists since $Z(G)$ is non-trivial. Write $C=C_G(A)$ and suppose $A \lneq C$. It is easy to show $C$ is normal in $G$. Now by standard $p$-group theory (see Lemma 1.23 in I.M. Isaacs, Finite Group Theory), we can find $B \unlhd G$, with $A \leq B \leq C$, and index$[B:A]=p$. Since $B \subseteq C$, we have $A \subseteq Z(B)$. Hence $B$ must be abelian (apply that if $|B/Z(B)| \text{ divides }p$, then $B$ is abelian). This contradicts the maximality of $A$ and the proof is finished.

Lemma 2 Let $G$ have a faithful primitive irreducible character, and let $A \lhd G$ be abelian. Then $A \subseteq Z(G)$.
Proof This follows from standard character theory and normal subgroups. Let $\chi \in Irr(G)$ be primitive and faithful. Primitiveness implies that $\chi_A=e\lambda$, for some positive integer $e$ and $\lambda \in Irr(A)$. Hence $A \subseteq Z(\chi)$, since $\lambda$ is linear. But $\chi$ is faithful, so $Z(\chi)=Z(G)$.

Now let us proceed with the proof. Let $\chi \in Irr(G)$ be primitive, $G$ being a $p$-group. Then $\chi$ can be viewed as an irreducible character of $G/ker(\chi)$. In this quotient group $\chi$ is both faithful and primitive. By combining Lemma 1 and Lemma 2, we see that $G/ker(\chi)$ must be abelian, which means that $\chi$ is a linear character indeed.

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  • $\begingroup$ Thanks. So you mean $\chi$ is induced by a linear character, and since $\chi$ is primitive, it has to actually be that linear character? $\endgroup$ – Kally Aug 12 '16 at 6:59
  • $\begingroup$ Exactly, you got it. $\endgroup$ – Nicky Hekster Aug 12 '16 at 8:27
  • $\begingroup$ I really appreciate the expanded answer! I know the proof that nilpotent groups are monomial, but it seemed to trivialize the problem. $\endgroup$ – Kally Aug 13 '16 at 3:02

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