Burnside before Burnside's transfer theorem As I already said, I am reading Part VIII of Burnside's "Notes on the Theory of Groups of Finite Order". (Proceedings of the London Mathematical Society, vol. XXVI, p. 325-338, 1895; The Collected Papers of William Burnside, vol. 1, p. 589-596.) It is the paper where Burnside proves that a finite non-abelian simple group with even order has order divisible by 12, by 16 or by 56.
From p. 591 of the Collected Papers, seventh line by the end, Burnsides assumes that
(hyp. 1) $G$ is a finite group whose Sylow 2-subgroups are of order 8;
(hyp. 2) if $P$ is a Sylow 2-subgroup of $G$, two elements of $P$ are conjugate in $G$ only if they are conjugate in $P$
and he announces that he will prove that G is not simple.
Note that if the Sylow 2-subgroups of $G$ are abelian, hyp. 2 is equivalent to say that for every Sylow 2-subgroup $P$ of $G$, $N_{G}(P) = C_{G}(P)$, thus, in this case, the non-simplicity of $G$ is today an easy consequence of Burnside's transfer theorem. But, in his 1895 paper, Burnside doesn't use his transfer theorem (I presume that he had not yet found it) and I would like to understand his reasonings. It could be a good little work on the history of mathematics.
At the start of his paper, Burnside proved that if the Sylow 2-subgroups of $G$ are cyclic, then $G$ is not simple. Thus, if we assume that the Sylow 2-subgroups of $G$ are abelian, we are left with two cases :
(i) the Sylow 2-subgroups of $G$ are isomorphic to $C_{2} \times C_{2} \times C_{2}$;
(ii) the Sylow 2-subgroups of $G$ are isomorphic to $C_{2} \times C_{4}$.
I understand Burnside's proof (that $G$ is not simple) for case (i), but not for case (ii).
Here is his proof (p. 593 in the Collected Papers, vol. I) for case (ii) :
"If next, the sub-groups of order $2^{3}$ are of of type (ii), and if the main group contains 3 distinct sets of conjugate operations of order 2, one of these sets contains exclusively operations which are the squares of operations od order 4, and the other two sets those that are not.
Let, now, $A$ be on operation of order 2 which is the square of an operation of order 4, and let $B$ an operation of order 2 belonging to a different conjugate set from $A$. Then $A$ and $B$ must generate a dihedral group of order $4n$, where $n$ is odd. Suppose [why "suppose" ? Isn't it necessarily true ?] that $AB$ is an operation of this group of order $2n$, and write
$(AB)^{n}= C$, $(AB)^{2}=S_{n}$,
so that $C$ is an operation of order 2, and $S_{n}$ an operation of order $n$. The operation $C$ must clearly belong to a different conjugate set from both $A$ and $B$. Now
$A S_{n} A = S_{n}^{-1}, B S_{n} B = S_{n}^{-1}, C S_{n} C = S_{n}$."
So far, so good. (Instead of "operation of a group", read "element of a group", and instead of "conjugate set of operations", read "conjugacy class of elements".) A complete proof of the preceding results is a bit long, but I can give it if anyone asks for it.
Then Burnsides says : "If $A^{1}$ is any operation contained in the sub-group within which the cyclical sub-group generated by $S_{n}$ is self-conjugate, and belonging to the same conjugate set as A [I understand : if $A^{1}$ is any $G$-conjugate of $A$ belonging to $N_{G}(<S_{n}>)$], then
$A^{1} S_{n} A^{1} = S_{n}^{-1}$,"
Why ? How do we know that we don't have, for example, $A^{1} S_{n} A^{1} = S_{n}$ ?
I copy the rest of Burnsides's proof : "and, therefore, $S_{n}$ cannot certainly be permutable [read : commute] with any operation of order 4, since it is not permutable with the square of any such operation. The operation $S_{n}$ therefore forms one of a set of $4r$ conjugate operations, where $r$ is odd; [I read : the number of $G$-conjugates of  $S_{n}$ is $4r$, where $r$ is odd;] and, when these are transformed by any operation of order 4, the resulting substitution [read : permutation] of the permutation-group consists of $r$ cycles of 4 symbols each. This is an odd substitution, and, therefore, again, in this case, the group cannot be simple."
I agree that $S_{n}$ commutes with no element of order 4 of $G$ (I can give a proof if anybody asks), but I don't see how Burnside can say that the number of $G$-conjugates of  $S_{n}$ is $4r$, where $r$ is odd. I obtain something different : [Edit : if $G$ is simple,] the number of $G$-conjugates of  $S_{n}$ is $\equiv 2 \pmod{4}$.
So my question is : do you think that Bunside's proof is correct and, if it is the case, could you explain it ? Thanks in advance.
 A: Let me turn my comments into an answer.
We have $A \in N_G(S_n) \setminus C_G(S_n)$ and $C \in C_G(S_n)$, so $|C_G(S_n)|$ and $|N_G(S_n)/C_G(S_n)|$ are both even.
We claim that $|C_G(N)|$ is not divisible by $4$. Suppose that it is. Then the subgroup $H := \langle C_G(S_n),A \rangle$ has order $2|C_G(S_n)|$, and so $H$ contains a Sylow $2$-subgroup of $G$.
By Sylow's Theorem, we can extend the subgroup $\langle A,C \rangle$ of $H$ to a Sylow $2$-subgroup $P$ of $H$ (with $|P|=8$). Since $|H:C_G(S_n)| = 2$, we have $|P:P \cap C_G(S_n)| = 2$. Let $Q = P \cap C_G(S_n)$ - so $|Q|=4$.
If $Q$ is cyclic, then $C$ is the square of an element of order $4$, but we have seen that $C$ is not conjugate to the element $A$ (which is a square), so this cannot happen. On the other hand, if $Q=C_2 \times C_2$ then, since $P$ is abelian, $P = C_2 \times C_2 \times C_2$, contrary to assumption. This establishes the claim.
So $C_G(P)$ has twice odd order, and all elements of order $2$ in $C_G(P)$ are conjugate to $C$. But $A^1$ is conjugate to $A$, not to $C$, so we cannot have $A^1 \in C_G(S_n)$.
In fact $A^1$ cannot centralize any  power $S_n^k$ with $k < n$  of $S_n$, or we could repeat the same argument with $S_n^k$ in place of $S_n$. So we must have $A^1S_nA^1 = S_n^{-1}$.
