1
$\begingroup$

I want to solve the following problem from Dummit & Foote's Abstract Algebra text (p. 186):

Let $H$ be a group of order $n$, let $K=\text{Aut}(H)$ and form $G=\text{Hol}(H)=H \rtimes K$ (where $\varphi$ is the identity homomorphism). Let $G$ act by left multiplication on the left cosets of $K$ in $G$ and let $\pi$ be the associated permutation representation $\pi:G \to S_n$.

(a) prove that the elements of $H$ are coset representatives for the left cosets of $K$ in $G$ and with this choice of coset representatives $\pi$ restricted to $H$ is the left regular representation of $H$.

(b) Prove $\pi(G)$ is the normalizer in $S_n$ of $\pi(H)$. Deduce that under the regular representation of any finite group $H$ of order $n$, the normalizer in $S_n$ of the image of $H$ is isomorphic to $\text{Hol}(H)$. [Show $|G|=|N_{S_n}(\pi(H))|$ using Exercises 1 and 2 above.]

(c) Deduce that the normalizer of the group generated by an $n$-cycle in $S_n$ is isomorphic to $\text{Hol}(Z_n)$ and has order $n \varphi(n)$.


My attempt:

(a) We know that there are $|G|/|K|=|H|=n$ left cosets of $K$ in $G$. If $h_1K=h_2K$ are the same coset, where $h_1,h_2 \in H$ then $h_1^{-1}h_2 \in K$. Since $h_1^{-1}h_2 \in H$ as well we find $h_1^{-1}h_2=1$ or $h_1=h_2$. Thus the $n$ distinct elements of $H$ give rise to $n$ distinct left cosets of $K$ in $G$, which are all such cosets.

Moreover, $\pi \big|_H(h)(h'K)=hh'K$, so that working with the representatives from $H$, $\pi \big|_H$ coincides with the left regular representation of $H$.

(b) I could prove one inclusion: Let $\pi(g) \in \pi(G)$. We have

$$\pi(g) \pi(H) \pi(g)^{-1}=\pi(gHg^{-1})=\pi(H),$$ hence $\pi(G) \leq N_{S_n}(\pi(H))$. I can't see how to use Exercises 1 and 2 (which I will quote below) to follow the hint.

(c) Once I have part $(b)$ right, this is easy.


My questions:

  1. Is my partial solution correct so far? If not, please help me fix it.

  2. How can I prove part (b) by following the hint in the brackets. For reference, Exercises 1 and 2 state that for a semi-direct product $G=H \rtimes_\varphi K$ we have $C_K(H)=\ker \varphi$ and $C_H(K)=N_H(K)$.

Thank you!

$\endgroup$
3
+100
$\begingroup$

Your answer to (a) looks OK to me.

Let $\Omega$ be the set of left cosets of $K$ in $G$. Let $N$ be the normalizer in $S_n$ of $\pi(H)$, and let $N_K$ be the stabilizer in $N$ of $K \in \Omega$.

Since $\pi(G) \le N$ (you have shown that), $N$ is transitive on $\Omega$, so $|N| = n|N_K|$ and $|\pi(G)|= |\pi(H)||\pi(K)|=n|\pi(K)|$, so it will be enough to prove that $|N_K| = |\pi(K)|$. But $\pi(K) \le N_K$, so we just need to prove that $|N_K| \le |\pi(K)|$.

We know that $\pi$ restricted to $H$ is equivalent to the left regular representation of $H$, so $\pi$ is injective when restricted to $H$. Hence we can identify $H$ with its image under $\pi$, and from now on we will write $H$ instead of $\pi(H)$.

Now the conjugation action of $N_K$ on $H$ induces a homomorphism $N_K \to {\rm Aut}(H)=K$, with kernel $C_{N_K}(H)$, so it will be enough to prove that $C_{N_K}(H) = 1$.

Let $g \in C_{N_K}(H)$. Now $g$ stabilizes $K$, so $g(K) = K$, and hence $$hK = h(K) = h(g(K)) = (hg)(K) = (gh)(K) = g(h(K)) = g(hK).$$ So $g$ fixes every coset $hK \in \Omega$ and hence $g$ is acting as the identity on $\Omega$. i.e. $g=1$ and $C_{N_K}(H)=1$ as required.

I am afraid that I have not used the hint involving Exercises 1 and 2, and I can't see immediately how to do that.

Incidentally, although this question seems to be for a finite group $H$, and the way I have worded the proof seems to assume this, the assumption is not really necessary. It is true also for infinite groups $H$.

$\endgroup$
  • 1
    $\begingroup$ Thank for the answer Derek. I can't really see how $N_K$ acts on $H$ by conjugation. I mean, the elements of $N_K$ are permutations $\sigma$ on the set $\Omega$ of left cosets (or equivalently, on the set $\{1,2,\dots,n\}$). While the elements $h \in H$ come from an arbitrary group. How can you make sense of an expression of the form $\sigma h \sigma^{-1}$? $\endgroup$ – user1337 Feb 12 '15 at 12:05
  • $\begingroup$ I mean the conjugation action of $N_K$ on $\pi(H)$ rather than on $H$. Or, equivalently, we can identify $H$ with its image under $\pi$. I have added a remark saying we will do that. $\endgroup$ – Derek Holt Feb 13 '15 at 15:12
  • $\begingroup$ @DerekHolt I know that this was a while ago, but could you please explain why your sentence "Since $\pi(G) \leq N$ [...] $=n|\pi(K)|$" is true? $\endgroup$ – CuriousKid7 Nov 30 '16 at 2:44
  • $\begingroup$ That is a complicated sentence. Could you make it clear exactly which assertion it is that you do not understand? If there are several, then what is the first such? $\endgroup$ – Derek Holt Nov 30 '16 at 10:37
  • 1
    $\begingroup$ $|N| = n|N_K|$ follows from the Orbit-Stabilizer Theorem. $|C_{N_K}(H)=1$ actually gives you $|N| \le |K|$, but $|K| = |\pi(K)|$. $\endgroup$ – Derek Holt Nov 30 '16 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.