2
$\begingroup$

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.

  • 1
    G can act on itself in different ways (by right or left multiplication, by conjugation ...) - does the question specify or imply the action which is intended?2012-05-04
  • 0
    It doesn't specify the action. I have been given the hint to consider the set A\U1 and the fact that the group generated by this set is normalized by G. But I don't see how this hint can help me.2012-05-04
  • 0
    That suggests that the action is by conjugation. Are the subgroups $U_1$ and $U_2$ supposed to be normal?2012-05-04
  • 0
    I'll ask my teacher if there's been a mistake because the problem does not specify the action. I only know that U1 and U2 are subgroups.2012-05-04
  • 0
    The action has to be by conjugation, since an action by left or right multiplication on a nonempty set would require the set to be the whole group.2012-05-04
  • 0
    Left and right multiplication can be ruled out but isn't there a lot of other possible actions than the three mentioned (and the trivial which wouldn't hold either)?2012-05-04
  • 0
    @M88: If the action were trivial, then $A$ would have to be a singleton for the action to be transitive, so $G$ would be cyclic with the single element of $A$ a generator; hence at least one of $U_1$ and $U_2$ contains a generator for $G$, and the conclusion would follow. Note that the action by conjugation is trivial when the group is abelian.2012-05-04
  • 0
    I think you misunderstood my question. Isn't there a bunch of different group actions on a subset? Just because we can rule out left/right multiplication and the trivial one we can't deduce that it must be by conjugation because there's a lot of other possibilities.2012-05-04
  • 0
    @M88: There are only a few *natural* actions, so only a few that could be "left unsaid" and have the problem make sense. The only reasonable ones that could be left unsaid are left and right multiplication, and conjugation. I think *you* misunderstood the question; the question doesn't ask you whether this hold for an *arbitrary* transitive action on the set $A$, the question assumes that the action is *understood from context*. So it should be conjugation.2012-05-04
  • 0
    The statement is in any case false for an arbitrary action. For example let $G=A_4$, and $A = \{(1,2,3),(1,3,2),(2,3,4),(2,4,3) \}$. Since $G$ does have a transitive action on 4 points, it is possible to define one on $A$. But there exist proper subgroups $U_1,U_2$ with $A < U_1 \cup U_2$.2012-05-04
  • 0
    You are right, the action is by conjugation. I haven't worked a lot with group actions so I didn't like the idea of restricting to one action unless the problem clearly states it. Thank you for your help so far.2012-05-04
  • 0
    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

2

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 $U_1=G$.