There is also a character-theoretic argument. Suppose $G' cap Z(G)$ has a subgroup $U$ of order $p$. We want a contradiction. Let $lambda$ be a nonprinciipal linear character of $U$. Since $U subseteq P$ and $P$ is abelian, $lambda$ has an extension to $mu$, a linear character of $P$. The induced character $mu^G$ has degree $|G:U|$, which is prime to $p$, so some irreducible constituent $chi$ of $mu^G$ has degree not divisible by $p$. Then $mu$ is a constituent of the restriction $chi_P$ by Frobenius reciprocity, and thus $lambda$ is a constituent of $chi_U$. But $U$ is central, so $chi_U = chi(1)lambda$. Now let $sigma$ be the linear character det$(chi)$. Then $sigma_U = lambda^{chi(1)}$, which is nontrivial since $p$ does not divide $chi(1)$. This is a contradiction, however, since $U subseteq G' subseteq {rm ker}(sigma)$. [Note that transfer proofs can often be replaced by arguments using the determinant of a character.]
No comments:
Post a Comment