0

The question comes from the top answer from Prove that there is no element of order $8$ in $SL(2,3)$

It says that "the claim follows by induction on $r$ (since a normal subgroup of order $m$ is a Hall subgroup and thus characteristic)."

What I try to continue through one of the following:

  1. Say at the end of induction chain we jump from $N \to M$, where $N$ has order $2m$ and $M$ has order $m$ and contain all even elements from $N$. I want to say that automorphism preserves parity, but doesn't that $S_6$ exception prevent us from doing so?
  2. I understand Hall subgroup's definition but do not know how to make a jump to the characteristic property?
Shaun
  • 47,747
SmoothKen
  • 439
  • Related but does not answer my question: https://math.stackexchange.com/questions/2495868/group-of-order-2n-m-m-odd-with-a-cyclic-sylow-2-subgroup-has-a-characte – SmoothKen Aug 20 '21 at 22:33
  • The outer automorphism of $S_n$ still preserves the subgroup $A_n$. For the second point, if there is a unique subgroup of a given order it must be characteristic, not just normal. – David A. Craven Aug 20 '21 at 22:36
  • Could you elaborate on your first statement? – SmoothKen Aug 20 '21 at 22:40
  • As $A_6$ is the unique subgroup of index $2$ in $S_6$, any automorphism of $S_6$ normalizes it. It swaps cycle types $3,1^3$ with $3^2$, and in $S_6$ it swaps $4,1,1$ with $4,2$, and $2,1^4$ with $2^3$, but it preserves parity. – David A. Craven Aug 20 '21 at 22:45
  • I see. Although one thing, $N$ is not $S_n$ and all we have is an injective map from $N \to S_n$. So even if we know all automorphism of $S_n$ preserve parity, does that somehow transfer to automorphisms of $N$? – SmoothKen Aug 20 '21 at 22:53
  • Parity just means: $G$ is a group and $H$ is a subgroup of index $2$ in $G$. Elements of $H$ are called 'even' and elements of $G$ are called 'odd'. If $G$ has a unique subgroup of index $2$, and $G$ is normal in an even larger group $L$, then elements of $L$ conjugate $H$ to a subgroup of index $2$ in $G$. The only choice is $H$. – David A. Craven Aug 20 '21 at 22:57
  • (Of course, elements of $G$ in that comment means elements of $G\setminus H$.) – David A. Craven Aug 21 '21 at 08:07
  • Yeah. It seems that this route will rely on the uniqueness property (as in the second route) anyway.... – SmoothKen Aug 21 '21 at 08:11

1 Answers1

1

A Hall subgroup of a finite group $G$ is a subgroup $H$ of order $m$ such that $m$ and $|G|/m$ are coprime.

Suppose that $H$ is a normal Hall subgroup of $G$ of order $m$. We claim that $H$ is the unique subgroup of order $m$ in $G$, which implies that it is characteristic.

To see this, suppose that $K$ is another subgroup with $|K|=m$. Since $H$ is normal in $G$, $HK$ is a subgroup of $G$ of order $|H||K|/|H \cap K| = m|K:H \cap K|$.

If $H \ne K$, then $|K:M \cap K| \ne 1$, and it divides $|K| = |H|$, but it also divides $|G:H|$, contradicting the coprimality of $m$ and $|G:H|$. So $H=K$ and $H$ is the unique subgroup of order $m$.

Derek Holt
  • 96,726