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Let $G$ be a subgroup of $S_n$, which acts transitively on $I= \{1, \ldots, n \}$. Let $N$ be a nontrivial normal subgroup of $G$. Then $N$ has no fixed points in $I$.

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Thanks, I'll remember. – P. M. O. Jun 2 '12 at 16:19
up vote 3 down vote accepted

Let $\,\,\{1\}\neq N\triangleleft G\,\,$ , and wlog let us assume $\,\,1\in\{1,2,...,n\}\,$ is a fixed point of $\,N\,$. But then for any $\,g\in G\,\,,\,x\in N$ , we get $$g^{-1}xg(1)=1\Longrightarrow xg(1)=g(1)\Longrightarrow g(1)\,\,\text{is a fixed point of } N$$

Well, now use that $\,G\,$ is transitive to get a contradiction

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Jeje...of course, thanx! I'll edit it – DonAntonio Jun 2 '12 at 16:32
Thanks don for the answer. – S. Snape Jun 2 '12 at 18:43
Then, as $G$ is transitive exists $g$ in $G$ such that $g (1) = j$, $j \in I$, right? – P. M. O. Jun 2 '12 at 20:18

Hint: If $x$ is a fixed point of $N$, then what is $\sigma x$ in relation to $\sigma N \sigma^{-1}$ for some $\sigma\in G$?

Now use $N$'s normality and the transitivity of the $G$-action...

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