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I have hard time understanding Fraleigh's proof. Can someone either explain Fraleigh's or provide an alternative, perhaps easier proof?

Thanks so much.

Math Damon
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    What is Fraleigh's proof? – Qiaochu Yuan Apr 01 '13 at 02:00
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    Have you looked at wikipedia? http://en.wikipedia.org/wiki/Cayley%27s_Theorem – Ian Coley Apr 01 '13 at 02:00
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    @MathDamon: I believe you are referring to "A First Course in Abstract Algebra, 7th Edition, John B. Fraleigh". You might also want to tell the readers which proof that is (number, page ...) and I think this is what Qiaochu Yuan is saying. Readers will not have a clue of what you speak. Even if they know the book, they may not have a copy. Regards – Amzoti Apr 01 '13 at 02:05
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    I don't know this Fraleigh's proof. But Cayley's theorem is easy and I don't think it admits many decent proofs. For each $g\in G$ you have a bijection $\sigma_g:g'\longmapsto gg'$ of $G$, your multiplicative group with unit $e$. The mapping $g\longmapsto \sigma_g$ is an injective group homomorphism (with left inverse $\sigma\longmapsto \sigma (e)$). That's your embedding of $G$ into $\rm{Sym} G$. – Julien Apr 01 '13 at 02:41

2 Answers2

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Let $n=|G|$. Think about the elements of $S_n$ - these are permutations of the numbers $1$ to $n$. Now let's arbitrarily number the elements of $G$ so that $G=\{g_1,g_2,\ldots, g_n\}$. Now, we could talk about the permutations in $S_n$ being applied to these elements of $G$, correct? For example, $(123)$ would permute $g_1$ and $g_2$ and $g_3$ accordingly. We call this "letting $S_n$ act on $G$."

Cayley's theorem does exactly what we just did, only backwards. He starts by labeling $G=\{g_1,g_2,\ldots, g_n\}$. Then he picks an element $x\in G$ and multiplies everything in $G$ on the left by $x$. Doing so rearranges the elements in $G$ in some way from their original order. He then finds the associated permutation in $S_n$, which he calls $\theta_x$.

Example. Let $|G|=3$ and suppose you start with elements $g_1,g_2,g_3$ (in that order). Then, suppose you multiply everything by some $x\in G$, and you end up with $g_2,g_3,g_1$ (in that order). Then $x$ will be associated with the permutation $\theta_x=(123)$ in $S_3$.

Once he finds this with all the $x\in G$, he looks at the subset $\{\theta_x:x\in G\}\subseteq S_n$. He then proves that that set is a subgroup, and in fact is isomorphic to $G$.

So, to sum everything up, Cayley's theorem simply shows that the permutations of a $G$'s elements induced by left multiplication by each element $x\in G$ form a group which itself isomorphic to $G$. Read the above explanation a few times and I think you'll see.

Alexander Gruber
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It's basically all about that the maps $g\cdot:G\to G$ (sending $x\mapsto gx$) are distinct permutations, that is, $(g\cdot)\in Sym(G)$, moreover this assignment $G\to Sym(G)$, sending $g\mapsto (g\cdot)$ is an injective group homomorphism.

Berci
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