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Let $G$ be a group with $25$ elements and $E$ a $G$-set with $32$ elements. Show that there exists $a \in E$ such that $G_a=G$.

So I want to show that $G_a=\{g \in G \mid ga = a\} = G$. I believe this $G_a$ is called the stabilizer of $G$?

I also found out that there is a theorem called the Orbit-Stalibizer theorem which states that there exists a bijection $\varphi:G/G_a \to Ga$ such that $gG_a \longmapsto ga$.

Can I use this to show the wanted result?

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1 Answer 1

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The orbits partition.

Since the stabilisers orders have to divide $25$ (they're subgroups), they're all of order $1,5$ or $25$.

So the same can be said for the orders of the orbits (orbit-stabilizer theorem).

But, there's no way to add up multiples of $1,5$ and $25$, and get $32$ without at least one $1$.

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  • $\begingroup$ So is the statement false? $\endgroup$
    – Walker
    Commented Aug 13, 2022 at 9:37
  • $\begingroup$ The statement is true. The orbit of size $1$ corresponds to a stabiliser of size $25$. $\endgroup$
    – i can try
    Commented Aug 13, 2022 at 15:42

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