TITLE: Exercise 5C10 in Isaacs' Finite Group Theory
QUESTION [6 upvotes]: Problem: Suppose that $G$ is simple group and has an abelian Sylow $2-$subgroup of order $8$. Show that the order of $G$ is divisible by $7$. 
Is there any hint to solve this problem? I'll be glad if one gives an answer.
Here's my start:
Let $|G| = 8k$ with $2\nmid k$. Then $n_2 = 1 \mod 2$, and $n\mid k$. Since $G$ is simple, $n_2 \neq 1$. After that I get stuck.

REPLY [8 votes]: Isaacs's Prop 5.18 states that whenever $G$ is a finite group with an abelian Sylow $p$-subgroup $P$, then $Z(N_G(P)) \cap G' \cap P = 1$. In our case $G=G'$ so we get that $Z(N_G(P)) \cap P = 1$. Of course $Z(N_G(P)) \cap P = C_P( N_G(P))$ are exactly those elements of $P$ that are left alone by every conjugation from $N_G(P)$.
Since the group of conjugations from $N_G(P)$ is exactly $N_G(P)/C_G(P) \leq \newcommand{\Aut}{\operatorname{Aut}}\Aut(P)$, and since $P \leq C_G(P)$ so that $N_G(P)/C_G(P)$ must be a group of odd order, we are interested in the odd order subgroups of $\Aut(P)$ for $P$ an abelian group of order 8.
If $P=C_8$ then $\Aut(P) \cong C_2 \times C_2$ has no non-identity subgroups of odd order, so $N_G(P)/C_G(P) = 1$ and $N_G(P) = C_G(P)$ and $C_P( N_G(P)) = P \neq 1$. Oops.
If $P=C_4 \times C_2$ then $\Aut(P) \cong D_8$ has no non-identity subgroups of odd order, so oops again.
If $P=C_2 \times C_2 \times C_2$ then $\Aut(P) \cong \operatorname{GL}(3,2)$ has odd order subgroups of orders 1, 3, 7, and 21. The ones of orders 1 and 3 centralize some non-identity elements of $P$, so oops. The ones of orders 7 and 21 are fine.
The one of order 7 creates what is called AGL(1,8) fusion and produces the simple group PSL(2,8). The one of order 21 creates what is called AΓL(1,8) fusion and produces the simple group J1 and ${}^2G_2(3^{2n+1})$ for $n \geq 1$.