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In Algebra by Michael Artin, Chapter 6, page 182 (second edition, Pearson), Proposition 6.11.2 states that there exists a bijective correspondence between operations of a group $G$ on the indices set $S=\{1,\cdots,n\}$ and permutation representations $G\to S_n$, in which a permutation representation of a group $G$ is a homomorphism from the group to a symmetric group, say, $\phi:G\to S_n$.

This is really driving me insane, because all I can imagine is that there is always solely one representation. My reasoning is straightforward, for each $g\in G$, it sends $1,\cdots,n$ to $g(1),\cdots,g(n)$, and thus corresponds to a $\sigma\in S_n$ such that $\sigma(1)=g(1),\cdots,\sigma(n)=g(n)$. So I think that the map $\phi:G\to S_n$ is completely determined by $G$ and there is thus but one representation.. Anything wrong?

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  • $\begingroup$ The map $\phi$ is completely determined... by the operation of $G$ on $S$ ! Your notation $g\left(1\right), \ldots, g\left(n\right)$ only makes sense with reference to an operation of $G$ on $S$. $\endgroup$ – darij grinberg Aug 12 '15 at 17:34
  • $\begingroup$ @darijgrinberg for each $g\in G$ I can do just the same thing, so doesn't $G$ completely determine the map? $\endgroup$ – Vim Aug 12 '15 at 17:36
  • $\begingroup$ @Vim: Just remove any mis-typed tag (from your post) and it will be purged automatically in due course (because of being an used "tag"). To clarify your Question, what is $n$? Is it the order of the group, $|G|$? I think that's what you meant to say, and it makes all the difference. $\endgroup$ – hardmath Aug 12 '15 at 17:37
  • $\begingroup$ @hardmath the author didn't seem to say anything about $|G|$, I think $n$ is just the number of elements of the indices set $S$. $\endgroup$ – Vim Aug 12 '15 at 17:39
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    $\begingroup$ @hardmath So perhaps I'm misunderstanding the meaning of a "group action", I thought it meant the action imposed by any $g\in G$ on the set $S$. Maybe the correct understanding should be a "law"? Say, a $*:G\times S\to S$ where the $*$ is an action? $\endgroup$ – Vim Aug 12 '15 at 17:57
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The thing that's wrong is that when you say $g(1),\ldots,g(n)$ you've already implicitly specified an action of $G$ on $\{1,\ldots,n\}$. So of course once you "know" what $g(1),\ldots,g(n)$ are, then there is only one action of $G$ on $\{1,\ldots n\}$ where $g$ acts by sending $i$ to $g(i)$.

Let's do a simple example. Let $G = \{\text{id},g\}$ be the cyclic group of order 2 (so that $g^2 = \text{id}$), and let $n = 3$, and lets look at ways that $G$ can act on the set $\{1,2,3\}$.

Of course $\text{id}$ must act trivially, on $\{1,2,3\}$, so to specify an action it suffices to determine how $g$ acts on $\{1,2,3\}$. Since $g$ has order 2, it must act as a permutation $\sigma\in S_n$ with $\sigma^2 = (1)$. Thus, $\sigma$ must either be the identity permutation (1), or a transposition (12),(13), or (23). These are all legitimate choices that define a group action of $G$ on $\{1,2,3\}$, so in this case there are 4 possible actions of $G$ on $\{1,2,3\}$, defined by sending 1 to the identity permutation, and $g$ to either $(1),(12),(13)$, or $(23)$.

I mean, if you just note that a permutation representation of $G$ is just a homomorphism $G\rightarrow S_n$, it should be obvious that in general there will be many such homomorphisms. In particular, $S_n$ has plenty of automorphisms (consider conjugation by elements of $S_n$), and so composing any homomorphism $G\rightarrow S_n$ by any automorphism of $S_n$ will generally give you a different homomorphism $G\rightarrow S_n$.

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  • $\begingroup$ Thank you, sir. This is much to learn. $\endgroup$ – Vim Aug 12 '15 at 18:01

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