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This is about problem 5B.1 page 157 in Isaacs' "Finite Group Theory" book. This chapter is definitely giving me trouble. The problem reads:

Let $G$ be a finite group, $P \in \operatorname{Syl}_p(G)$ and suppose that $g \in P$ has order $p$, where p is a prime. If $g \in G'$, but $g \notin P'$, show that $g^t \in P'$ for some element $t \in G-P$.

Right. So, take $v:G \rightarrow P/P'$ to be the transfer homomorphism. Since $g \in G' \leq \operatorname{Ker}(v)$, we see that $V(g) \in P'$. Using the transfer-evaluation lemma, $V(g) = \prod_{t \in T_0}t^{-1}g^{n_t}t$. However, each $n_t$ divides $o(g)=p$. Hence $V(g) = \prod_{t \in A}t^{-1}gt \in P'$, since in the complement of $A$ in $T_0$ the expression vanishes. Also, again by the transfer-evaluation lemma, $t^{-1}gt \in P$ for all $t \in A$. I don't see how from these assumptions one can deduce the existence of a conjugate of $g$ in $P'$. Another thing to notice is that $|A| + p|T_0 - A|=\sum{n_t}=|G:P|$. In particular, this implies that $p$ does not divide $|A|$. So, we have a product of, $k$ say, elements of $P$ which lies in $P'$. If it were true that $V(g)=x^k$ for some $x=g^s \in P$, then, since $\gcd(k, |P:P'|)=1$, we could deduce that $x \in P'$ as required. A choice which could make sense would be to take the product of all elements in $A$ as $s$, but I don't see why $V(g^s)=g^{ks}$ should hold in that case. Of course, there might be a different argument that would work.

Any thoughts?

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@StefanosA: Let X be GL(3,3), P be one of its Sylow 3-subgroups, and G be the normalizer of P in X. Then every non-identity element of P has order p and is contained in G′, but the derived subgroup of P is normal in G, and so nothing gets conjugated into it. –  Jack Schmidt Sep 26 '11 at 10:26
Doing a brief search, perhaps what Isaacs had in mind was proposition 37.4 page 199 of Aschbacher's "Finite Group Theory" in the contrapositive form. –  the_fox Sep 26 '11 at 10:38
Sure. Let $G$ be the group $GL(3,3)$, and then let $P$ be the upper-triangular matrices with all 1's on the diagonal. Then $P\le G'$, and yet $P'$ is just the subgroup of $P$ where the first super-diagonal is all 0's. Now there are plenty of elements that work: for example, let $g$ be the matrix with all 1's on and above the diagonal. –  user641 Sep 26 '11 at 10:42
It makes sense to assume that what the author intended to write was $P$ instead of $P'$. –  the_fox Sep 28 '11 at 8:11
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