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About holomorph of a finite group being the normalizer of regular image

Mathematics Asked on January 7, 2022

Here is part of Exercise 5.5.19 in Dummit & Foote’s Abstract Algebra:

Let $H$ be a group of order $n$, let $K=operatorname{Aut}(H)$ and $G=operatorname{Hol}(H)=Hrtimes K$ (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 $pi:Gto S_n$.

(a) Prove 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 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 $operatorname{Hol}(H)$. [Show $|G|=|N_{S_n}(pi(H))|$.]

I could easily show (a), but even before attempting to prove part (b), I was puzzled: Does it imply that $pi$ is injective? The restriction of $pi$ to $H$ is injective by part (a), but I don’t think $pi$ is injective in general.
For example, $D_8simeqoperatorname{Aut}(D_8)$, so if $H=D_8$, then $K$ is a normal subgroup of $G$ (being of index 2), so $kerpi=Kneq1$.
Am I wrong somewhere in my deduction?

One Answer

In the case of $H=D_8$, it's not true $K$ is index $2$; actually $K$ is index $|H|$ in $G=Hrtimes K$. And $K$ is never normal in $G$, unless it's trivial of course.

The best way to think about $G$ is as an "affine group" of $H$. Indeed, if $H=mathbb{Z}_p^n$ then $G$ is literally the affine group of $H$ as a vector space. In general, we can think of $G$ as the subset of $S_H$ comprised of "affine functions" of the form $xmapsto alpha(x)b$ where $alphainmathrm{Aut}(H)$ and $bin H$.

Conjugating $K$ by $H$ yields functions of the form $alpha(xb^{-1})b=alpha(x)alpha(b)^{-1}b$; to be an automorphism of $H$ (element of $K$) it must preserve $ein H$ as a function, which requires $alpha(b)=b$ (which, conversely, is sufficient), and this is only true for all $b$ if $alpha$ is the identity automorphism. On the other hand, conjugating $H$ by $K$ yields functions $alpha(alpha^{-1}(x)b)=xalpha(b)$, which are still elements of $H$, so $H$ is normal in $G$.

Note $H$ is a transversal for $G/K$, and indeed $G$ acts on $H$ representing $G/K$ matches $G$ acting on $H$ by affine functions in the way I described above. You want to show $N_{S_H}(H)=G$ here.

Answered by runway44 on January 7, 2022

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