Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to solve exercise 8B.6 on page 249 of Isaacs's Finite Group Theory textbook (the second question in a series; this is the third as question here). I have an idea, but it doesn't quite work. It seems like the precursor to the idea in the next exercise, so I'd really like for it to work.

Can my idea be fixed?

The exercise asks:

If $G$ is a finite primitive permutation group with a point stabilizer $H$ that has an orbit of size 2, then $G$ is dihedral of order twice an odd prime in its natural action.

The previous exercise had the stabilizer's orbit size reduced to 1, and it followed by showing a maximal subgroup contained in two point stabilizers had to be the identity (which caused a huge collapse: $G$ is then cyclic of prime order). Ok, so I want to do the same thing, $H$ has a big subgroup contained in two stabilizers, so that big subgroup is the identity (which finishes the whole thing for my choice of subgroup).

Here was the smaller question I had while doing that, but I answered it depressingly as: "No."

Suppose $H$ is a maximal subgroup of the finite group $G$, and let $K = H^2 = E^2(H)$ be the subgroup of $H$ generated by the squares of the elements in $H$. If $g\in G$ takes $K$ to $K^g ≤ H$, then must $g$ normalize $K$?

Well, taking $G$ to be $\operatorname{PGL}(2,q)$ for $q ≡ ±3\bmod 8$, and $H$ to be a Sylow 2-subgroup (so a dihedral group of order 8 with partial fusion), then $K$ is the order 2 normal subgroup of $H$ and is not weakly closed: there is a $g\in G$ that takes $K$ to a non-normal subgroup of order 2 in $H$. When $q = 3$, $H$ is a maximal subgroup of $G ≅ S_4$.

Now the action of $G$ on the cosets of $P$ is primitive, but not faithful, so this isn't actually a counterexample, but wow it seems close.

Is it true that $K$ is weakly closed in $H$ (with respect to $G$) if $H$ is core-free (with respect to $G$)?

The only reason $H$ had a core was because $q$ was chosen so small. But all $q$ have the same basic setup as far as the relationship of $K$ in $H$ and conjugation in $G$, it is just that $H$ is no longer maximal for larger $q$.

share|improve this question
If the action of $G$ on the cosets of $P$ is primitive and faithful, then this is true, as $K$ is weakly closed in $P$ iff $K$ is normal in $N_G(P)$ (which is either $P$ or $G$ in this case). –  user641 Oct 21 '11 at 18:45
I would also like to say +1!, because when I (finally) worked these three exercises out, there seemed to be no underlying, unifying idea behind my ad-hoc techniques. –  user641 Oct 21 '11 at 18:48
Isn't this similar to the one with orbit size 3? Let $H=G_{\alpha,\beta}$. Then $|G_\alpha:H|=2$, hence also $|G_\beta:H|=2$, so $N_G(H)$ strictly contains $G_\alpha$, hence $H \lhd G$ and so $H=1$ and $|G_\alpha|=2$. –  Derek Holt Oct 22 '11 at 14:32
@SteveD: I think it is not true that K is weakly closed in P iff K is normal in N_G(P). For instance, when G=PGL(2,q) [q=±3 mod 8], Z(P) is normal in N_G(P), but there is a G-conjugate of Z(P) that is not normal in P. (exercise 5C.6 is strangely worded). –  Jack Schmidt Oct 22 '11 at 15:16
@Derek: Thanks. I was trying to use an element that took a to b to be my extra member of the normalizer. Just using G_b is a much better idea. –  Jack Schmidt Oct 22 '11 at 15:17
show 1 more comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.