# The normaliser of the left regular image [D&F]

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:

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!

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 $H$.

• 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 $\sigma h \sigma^{-1}$? – user1337 Feb 12 '15 at 12:05
• 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. – Derek Holt Feb 13 '15 at 15:12
• @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? – CuriousKid7 Nov 30 '16 at 2:44
• 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? – Derek Holt Nov 30 '16 at 10:37
• $|N| = n|N_K|$ follows from the Orbit-Stabilizer Theorem. $|C_{N_K}(H)=1$ actually gives you $|N| \le |K|$, but $|K| = |\pi(K)|$. – Derek Holt Nov 30 '16 at 22:39