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I am having some trouble understanding a part of the proof of the second theorem (the one after Cayley's theorem) in the handout on https://math.la.asu.edu/~kawski/classes/mat444/handouts/strongCayley.pdf.

It says that "The action of $G$ by left multiplication on $\mathcal{L}_H$ induces a map $\Phi:G\to S_{\mathcal{L}_H}\cong S_m$ via $\Phi(g)(aH)=(ga)H$."

But what does that really mean? Since the codomain of $\Phi$ is a symmetric group, I assume that $\Phi(g)$ is a permutation, acting on $aH$ in this case. If that is the case, then why should it equal $(ga)H$? Isn't that just another left coset of $H$, possibly not equal to $aH$?

I would be grateful if someone could point out where I am mistaken here.

And another short question: Is $S_{\mathcal{L}_H}$ a legit way of referring to a symmetric group? I thought the indices could only be positive integers?

mss
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1 Answers1

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For a set $A$ we have

$$S_A:=\{f:A\to A\ |\ f\mbox{ is a bijection}\}$$

by definition. Now if $n\in\mathbb{N}$ then $S_n$ is a shortcut for $S_{\mathbb{N}_n}$ where $\mathbb{N}_n:=\{1,2,\ldots, n-1, n\}$.

Now note that if $A$ is finite then $S_A\simeq S_{m}$ for $m=|A|$. The isomorphism is induced by any bijection $A\to\mathbb{N}_{m}$.

With that it is clear that $\Phi(g)$ is supposed to be a bijection on $\mathcal{L}_H$, i.e. $\Phi(g):\mathcal{L}_H\to\mathcal{L}_H$. And the definition is given by $\Phi(g)(aH)=(ga)H$.

freakish
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