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Let $G$ be a finite group acting on a finite set $A$. The action is said to be transitive if for every $a, b \in A$, there exists $g \in G$ such that $ga = b$. Prove that if $|A| > 1$ and the action is transitive, then there exists $g \in G$ such that $ga \neq a$ for every $a \in A$.

My attempt:

By way of contradiction, suppose that for every $g \in G$, it holds that $ga = a$, which would be the trivial action, and thus the transitive action does not hold, leading to a contradiction.

My question is this correct?

Shaun
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  • No. Transitive just means that, given $a,b\in A$, we can find some element $g_1\in G$ such that $g_1a=b$. Doesn't mean there can't be another element which fixes $a$. Nor does it mean that $g_1$ can't fix some elements of $A$. – lulu Oct 13 '24 at 17:02
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    From the [tag:solution-verification] tag wiki: "A question with this tag should include an explanation for why the argument presented is not convincing enough." – Shaun Oct 13 '24 at 19:30
  • The negation of "there exists $g\in G$ such that for all $a\in A$, $ga\neq A$" is not "for every $g\in G$ for every $a\in A$, $ga=a$". The negation is "for each $g\in G$ there exists at least one $a\in A$ such that $ga=a$." – Arturo Magidin Oct 13 '24 at 19:43
  • Arturo's right. You messed up your quantifiers, so to speak. So is Lulu (I think). I'd type an answer but sounds like this might be marked as a dupe. Nah, maybe I will anyway. – suckling pig Oct 13 '24 at 23:54
  • Hint: see Burnside's lemma. – suckling pig Oct 14 '24 at 00:27

2 Answers2

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The correct contradictory assumption is: $$\forall g\in G, \exists a\in A\mid ga=a$$ This means that $\left|\operatorname{Fix}(g)\right|\ge1$ for every $g\in G$. Therefore: $$\sum_{g\in G}\left|\operatorname{Fix}(g)\right|=\left|\operatorname{Fix}(e)\right|+\sum_{g\in G\setminus\{e\}}\left|\operatorname{Fix}(g)\right|\ge |A|+|G|-1$$ and hence (Burnside's counting lemma): $$\text{ number of orbits}\ge1+\frac{|A|-1}{|G|}>1$$ because $|A|>1$: contradiction.

suckling pig
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Kan't
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It happens you are incorrect.

As we know the group action gives us a homomorphism from $G$ into $S_A.$ The latter is the group of permutations of the set $A.$

In this parlance, the element $g$ we are looking for corresponds to a permutation is a derangement. It moves everything, or has no fixed point.

I'll try to help you prove it if you get stuck.

suckling pig
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