By $A_6$ I mean the alternating group on $6$ letters, which is not solvable and so needn’t have Carter subgroups, but if they exist will still be conjugate. Since the sylow $2$-subgroups are conjugate, nilpotent and to me seem to be selfnormalizing, I’m wondering if this is true, and if hence the sylow $p$-subgroups for $p=3$ and $5$ cannot be selfnormalizing? As stated here and on Wikipedia, $A_5$ has no Carter subgroups. What is the situation for An when $n>6$? For one thing, since the Sylow $2$-subgroups in $A_6$ are selfnormalizing, wouldn’t they necessarily remain that way for larger $n$ by conjugating a part of the sylow $2$-subgroup that is in a part of the $A_n$ that has just $6$ letters?(And they would remain conjugate and nilpotent).
Are the Sylow $2$-subgroups of $A_6$ Carter subgroups
group-theory
Related Solutions
I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:
Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.
Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.
Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.
Something like this ought to work for $A_5$.
EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.
Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.
The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.
Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.
Personally, I don't believe that the theory of central series is more elementary than Sylow theory, and in my experience students find Sylow theory easier, probably because it does not involve technical calculations with quotient groups.
But if you insist, I think you can do it this way. By using induction on $|G|$, we can assume that $Z(G)=1$. Also all maximal subgroups $M$ of $G$ are normal, so they have prime index $p$. It is not hard to show that $M$ satisfies the normalizer condition, so $M$ is nilpotent by induction. So $Z(M) \ne 1$. Let $N$ be a minimal normal subgroup of $G$ that is contained in $Z(M)$, let $G = \langle M,g \rangle$ with $g^p \in M$, and let $H = \langle N,g \rangle$.
So $Z(H) \cap N \le Z(G)$ and hence $Z(H) \cap N = 1$. Since $\langle g \rangle \cap N \le Z(H) \cap N$, we have $\langle g \rangle \cap N =1$.
Now $H$ satifies the normalizer condition, by assumption if $H=G$, and by induction otherwise. So $\langle g \rangle < N_H(\langle g \rangle)$, and hence $N_N(\langle g \rangle) \ne 1$. But $[N_N(\langle g \rangle), \langle g \rangle] \le \langle g \rangle \cap N =1$, so $N_N(\langle g \rangle) \le Z(G)$, contradiction.
Best Answer
The Sylow-2 subgroups of $A_6$ are self-normalizing because there are 45 of them, and the order of $A_6$ is 8*45.
There are 45 because each is an index two subgroup of the $2\times D_8$ Sylows in $S_6$. There are 15 ways to partition 6 points into 4 and 2 points, and, given that, 3 ways the $D_8$ can act on the four points, and these differences are preserved in $A_6$...e.g., one can have $(12)(3456)$, $(12)(3546)$, or $(12)(3465)$, in the Sylow, but these are all in distinct Sylows.
Thus a Sylow-2 in $A_6$ is a Carter subgroup.