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I'm trying to prove that if $G$ is a simple group and $H$ is a proper subgroup of $G$ of index $n$ then $G$ is isomorphic to a subgroup of $S_n$.

To do this, I've been trying to find a homomorphism between $G$ and permutations on the set of right cosets of $H$, because this would give me a subgroup of $S_n$, but I've failed to find one with a trivial kernel(or at least one I can prove has a trivial kernel). Nor am I sure how to use the condition that $G$ is simple. I'm not even convinced this is the right approach, but I can't think of another way to get permutations on $n$ things.

Thanks!

[Edited the title]

Shaun
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Kate
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1 Answers1

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Define $\phi_g(hH) = (gh)H$.

Show that this is independent of choice of h - if $h'\in hH$, then $(gh')H = (gh)H$.

Show that $\phi_{gg'}=\phi_{g}\circ\phi_{g'}$ and $\phi_e=id$, where $e\in G$ is the identity, and $id$ is the identity function on the set of left cosets of $H$.

Thus $g\rightarrow \phi_g$ is a homomorphism from $G$ to the set of permutations of the left cosets. There are $n$ left co-sets, to we have a homomorphism $G\rightarrow S_n$.

What can be said about the kernel of the homomorphism, given that $G$ is simple?

Thomas Andrews
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  • I don't understand the first line. Do you mean to take a fixed, n-tuple $(a_1H, a_2H,\dots, a_nH)$, which are all those cosets of $H$, and they are distinct. Then define $\phi(g)=(ga_1H, ga_2H,\dots, ga_nH)$? If this understand is correct, or equivalent to your definition, then we can prove $\phi(g_1g_2)=\phi(g_1)\circ\phi(g_2)$, so it is a homomorphism from $G$ to the permutation group of $n$ objects (cosets). So $\ker\phi$ is normal in $G$, and $G$ is simple, so $\ker\phi={e}$, finally, by 1st isomorphism theorem, the proof is done. –  May 16 '23 at 22:59
  • Where did anybody mention anything about $n$-tuples? The set of left cosets is a set, of size $n,$ and $G$ acts on the set. @GGplay – Thomas Andrews May 17 '23 at 18:10
  • Yes, but can I interpret it as a n-tuple? so each time the element of $G$ acts on this n-tuple, it will give a permutation of this n-tuple. As I wrote in the comment, can I prove it in this way? –  May 18 '23 at 03:18
  • It's a permutation of a set. Ou don't need tuples to understand permutations. Given a set $X,$ you a permutation of $X$ is a bijection $\pi:X\to X.$ There is no need for ordered tuples, that's just a way we sometimes notate permutation when $X={1,\dots,n}.$ But it is a very limiting notation, because it assumes there is a natural orser to $X.$ And if you question was "can I take this as...," don't phrase it as, "do you mean..." If you must interpret permutations this way, I suppose you can, but it is a limiting way to think of permutations. @GGplay – Thomas Andrews May 18 '23 at 19:23
  • When I learn the symmetric group $S_n$, the textbook uses the example like $\sigma=\binom{1234}{2413}$. It reads as from top to bottom as $1\to 2, 2\to4, 3\to1, 4\to 3$. and this gives me a "limited" feeling how the permutation works on a set. I interpret it as two rows, top and bottom, each row is like a 4-tuple in this textbook's example. So when I read your proof, I try to interpret it as the n-tuples. I have a question that is that always true that I can convert your abstract definition of permutations into this representations, such as $\sigma=\binom{12\dots n}{24\dots3}$? –  May 18 '23 at 22:35
  • Sorry, the context exceeds the word limit. Continued with above, in this notation, let $(a_1H, a_2H,\dots, a_nH)$ be all distinct cosets of $H$. I can define the map $\phi: G\to S_n$ as $$\phi(g)=\sigma_g=\binom{a_1H, a_2H,\dots, a_nH}{ga_1H,~~ ga_2H,\dots, ga_nH}$$ –  May 18 '23 at 22:43