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I've been struggling with this problem for some time now so any help would be greatly appreciated.

I have a finite group $G$ and a subset $A$ of $G$. I am told that $G$ acts transitively on $A$ and that $A$ generates $G$.

Furthermore I have two subgroups of $G$ called $U_1$ and $U_2$ such that $A$ is contained in $U_1\cup U_2$. I now have to prove that either $U_1=G$ or $U_2=G$.

What I've done so far:

I've tried solving it by contradiction so I assume that $U_1$ and $U_2$ are different from $G$. I found it fairly easy to say a few things about $U_1$ and $U_2$ such that they aren't equal, one is not contained in the other and their union isn't $G$ but I don't know what to do about the group action part.

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    It's a difficult problem even when you know what the action is! They could have made things much more transparent by not mentioning actions at all, and just saying that $A$ is a conjugacy class.2012-05-05

1 Answers 1

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Let's assume the action is conjugation and see if we can solve the problem. So $A$ is a conjugacy class of $G$.

Let $B = A \cap U_1$ and $C = A \setminus B$. So $C < U_2$. Let $H = \langle C \rangle \le U_2$.

For $g \in U_1$ and $c \in C$, we cannot have $g^{-1}cg \in U_1$, because that would imply $c \in U_1$. So $g^{-1}cg \in C$ and hence $U_1 \le N_G(H)$.

Since $H \le N_G(H)$ and $A \le U_1 \cup H$, every element of $A$ normalizes $H$. But $A$ generates $G$, so $H$ is normal in $G$.

Then if $C$ is nonempty, we have $A \le C^G \le H \le U_2$, so $U_2 = G$, whereas if $C$ is empty then $A \le U_1$ so $U_1=G$.