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Must an homomorphism induced by a transitive action be surjective?


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up vote 5 down vote accepted

No. If $X=\{1,\dots,n\}$ for some positive integer $n$, then the cyclic group $C_n$ of order $n$ with generator $g\in C_n$ acts on $X$ by the rule $g\cdot i=i+1$ for all $1\leq i\leq n-1$ and $g\cdot n = 1$. The action of $C_n$ on $X$ is transitive. However, $C_n$ is a proper subgroup of the permutation group of the set $X$.

The following supplementary exercises are relevant:

Exercise 1: Prove that if $H\subseteq C_n$ is a proper subgroup, then $H$ does not act transitively on $X$ in the context of the example above.

Exercise 2: Give an example of a subgroup $G\subseteq S_n$ minimal with respect to the property that $G$ acts $2$-transitively on $X$ in the context of the example above. (Please read Ben Millwood's answer below for the definition of a $2$-transitive action.)

I hope this helps!

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Thank you very much. I have 2 questions though: 1. Why does this rule define a group action? We have $1(i)=i+1 \neq i$ 2. Let us take $1,3 \in \{1,...,n \} $, then there is no $g \in G $ so that $ g(1)=3 $ – Roy Sep 1 '12 at 9:08
@Roy: sure there is, $g^2$. – Ben Millwood Sep 1 '12 at 9:10
maybe I misunderstood the action. You mean that $g^k(i)=i+k$? – Roy Sep 1 '12 at 9:13
Well, if $g(i) = i + 1$ we really don't have any choice about what $g^k$ does. – Ben Millwood Sep 1 '12 at 9:16
@Roy Do you understand the example now? I think Ben Millwood has explained the example in the comments above but if you need further explanation, then please do not hesitate to ask. I have added two exercises to my answer that might be useful; I think that you will gain a deeper understanding of this example if you think about them. – Amitesh Datta Sep 2 '12 at 2:34

There exists a concept called an $n$-transitive action, that is, an action for which any $n$-element subset of the set acted upon can be mapped to any other such set (so for example I can map any unordered pair to any other. Note that the definition uses subsets rather than tuples because I can never map two things to the same place, or one thing to two places).

Once you're aware of this concept and examples of it (e.g. Möbius transformations are triply-transitive on the Riemann sphere), the following two facts ought to become clear:

  • If the set acted on has $n$ elements, the induced homomorphism is surjective exactly when the action is $n$-transitive (since $S_n$ itself is clearly $n$-transitive).
  • There are certainly actions which are $1$-transitive or $2$-transitive or $k$-transitive but not anything more.

These two facts together show that homomorphism is not always surjective.

Another way of looking at it is to realise that if an action on $\{1\dots 6\}$ is transitive, that means I can map $1$ to $3$ and I can map $5$ to $4$ and I can map $2$ to $6$, but if the homomorphism is surjective, that means I can do all three with a single element of the group. That's clearly a much stronger condition.

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Thank you, the last note made it much clearer. – Roy Sep 1 '12 at 9:42

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