Exercise 5C10 in Isaacs' Finite Group Theory 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.
 A: 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 $n \geq 1$.
A: Solution: Let $P$ be an abelian Sylow $2−$subgroup of the simple group $G$. If a Sylow $2−$subgroup of a simple group is abelian, then it must be elementary abelian. So, $P \cong {Z_2} \times {Z_2} \times {Z_2}$. Recall that for the elementary abelian group $G$ of order ${p^n}$, 
$Aut\left( G \right) \cong GL\left( {n,p} \right)$. Note that 
$\left| {GL\left( {n,p} \right)} \right| = \prod\limits_{i = 0}^{n - 1} {\left( {{p^n} - {p^i}} \right)} $. So, 
$\left| {Aut\left( P \right)} \right| = 7.6.4 = 168$. Also note that 
$\frac{{{N_G}\left( P \right)}}{{{C_G}\left( P \right)}} \cong Aut\left( P \right)$
$ \Rightarrow $
$\left| {{N_G}\left( P \right)} \right| = \left| {{C_G}\left( P \right)} \right|.\left| {Aut\left( P \right)} \right| = \left| {{C_G}\left( P \right)} \right|.168$. Hence, 
$\left. 7 \right|\left| {{N_G}\left( P \right)} \right|$. By Langrange, we are done.
