# 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?

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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