# Thread: If |G|=42, and the 3-Sylow is not normal, for all elements s.t. |x|=3, |y|=7, xy!=yx

1. ## If |G|=42, and the 3-Sylow is not normal, for all elements s.t. |x|=3, |y|=7, xy!=yx

Here's what I've thought thus far.

There is only one 7-Sylow (easy to check). Then, the number of 3-Sylows is 7. We know that they are conjugates of each other. Let P be the 3-Sylow and Q be the 7-Sylow. Then consider Q acting on P via conjugation. The orbit stabilizer theorem implies that

$|\text{orbit}| = \frac{|Q|}{|\text{stabilizer}|}$

It follows that the orbit must divide 7, so either 1 or 7. If the orbit is 7, we are in luck: Conjugating by Q gives us all possible 3-Sylows and each of them is disjoint apart from the identity, so that means elements of the 3-Sylows can't possibly commute with elements of the 7-Sylow.

I have no idea what to do in the case where the orbit is just 1 element: i.e. the 7-Sylow is in the normalizer of the 3-Sylow. It doesn't imply that the 3-Sylow is normal, so there is no contradiction there. What more must I do?

2. Originally Posted by eeyore
Here's what I've thought thus far.

There is only one 7-Sylow (easy to check). Then, the number of 3-Sylows is 7. We know that they are conjugates of each other. Let P be the 3-Sylow and Q be the 7-Sylow. Then consider Q acting on P via conjugation. The orbit stabilizer theorem implies that

$|\text{orbit}| = \frac{|Q|}{|\text{stabilizer}|}$

It follows that the orbit must divide 7, so either 1 or 7. If the orbit is 7, we are in luck: Conjugating by Q gives us all possible 3-Sylows and each of them is disjoint apart from the identity, so that means elements of the 3-Sylows can't possibly commute with elements of the 7-Sylow.

I have no idea what to do in the case where the orbit is just 1 element: i.e. the 7-Sylow is in the normalizer of the 3-Sylow. It doesn't imply that the 3-Sylow is normal, so there is no contradiction there. What more must I do?

Since there are 7 Sylow 3-sbgps. of G, this means that the normalizer of any of them is a sbgp. of index 7

in G, which means the normalizer of any Sylow 3-sbgp. of G has order $\frac{42}{7}=6$ .

In particular, no element of order 7 fixes an element of order 3 and we're done...

Tonio