# Set of conjugates of element cannot be contained in the union of two proper subgroups

Edit: Using the excellent hints provided by @SMM, I was able to solve the problem. See my answer below.

I've been thinking off and on about this problem over the last couple of days and would appreciate a hint in the right direction. Please don't give me a full solution - I'll post one myself once I have one.

The problem is from Kurzweil and Stellmacher's Theory of Finite Groups, section 1.2 ("Homomorphisms and Normal Subgroups"). In this section, the isomorphism theorems and normal subgroups are discussed. But nothing much more sophisticated (e.g. group actions) has been covered yet. In this section, groups are not assumed to be finite unless stated otherwise.

Let $x \in G$, $D := \{x^g \ | \ g \in G\}$, and $U_i \leq G$ for $i = 1,2$. Suppose that $\langle D \rangle = G$ and $D \subseteq U_1 \cup U_2$. Then $U_1 = G$ or $U_2 = G$.
Paraphrasing this in my own words: if $G$ is generated by the set $D$ of conjugates of an element $D$, then at least three proper subgroups are required in order to "cover" all of $D$.

I haven't been able to solve this yet, but I have some ideas:

First, note that no element of $D$ can be contained in a proper normal subgroup, say $N < G$. If this were so, then all of $D$ must be contained in $N$. But then $G = \langle D \rangle \leq N < G$, which is absurd.

Now let's look at a few specific cases.

First, consider $|D| = 1$, then $D = \{x\}$, so $\langle x \rangle = G$, which means $G$ is cyclic. Then (say) $x \in U_i$, so $G = \langle x \rangle \leq U_i$, and therefore $U_i = G$ as desired.

Now consider $|D| = 2$. Here I'm bringing in some machinery which has not been covered yet, just so I can make some progress. If $G$ acts on $D$ by conjugation, then the stabilizer of $x$ is $\{g \in G \ | \ gxg^{-1} = x\} = \{g \in G \ | \ gx = xg\} = C_G(x)$, the centralizer of $x$. Moreover, the index $[G:C_G(x)]$ is equal to $|D| = 2$. Therefore, $C_G(x)$ is a proper normal subgroup of $G$ which contains $x$, which is disallowed as argued above.

So if $|D| \neq 1$ then $|D| \geq 3$, which seems promising since we're trying to rule out $D$ being contained in the union of two proper subgroups of $G$.

Unfortunately, it's not clear how I should proceed. Also, I'm concerned that what I have done so far is not in the spirit of what was intended. I think I should be formulating an argument involving normal subgroups and quotient groups. But there aren't any obvious nontrivial normal subgroups in sight.

The kernel of any homomorphism is normal, but the only obvious homomorphism that comes to mind in this problem is conjugation by $g$, which is a bijection, hence the kernel is trivial.

The intersection of all of the conjugates of any subgroup (say, one of the $U_i$) is normal, but I don't see anything that would ensure that this intersection is not trivial.

By the way, note that $G$ and $D$ do not appear to be assumed finite for this problem, so that's an extra wrinkle. But even a hint that only works in the finite case would be helpful.

It is sufficient to prove that $D\subseteq U_1$ or $D\subseteq U_2$. So, assume that it is not the case. Let $D_1= (D\cap U_1)\smallsetminus U_2$ and $D_2= (D\cap U_2)\smallsetminus U_1$. Then:

1. prove that $U_2\leq\mathrm N_G(\langle D_1\rangle)$ (denote $N=\mathrm N_G(\langle D_1\rangle)$);

2. choose $y\in D_2$ and prove that it is not conjugated in $N$ to any element of $D_1$; conclude that $N\lneq G$;

3. finally, note that $N$ contains $D$, hence $N=G$.

Using the excellent hints provided by @SMM, I was able to solve the problem.

First, we partition $D$ into three disjoint subsets: \begin{align} D_1 &= (D \cap U_1) \setminus U_2 \\ D_2 &= (D \cap U_2) \setminus U_1 \\ D_{12} &= D \cap U_1 \cap U_2 \\ \end{align} If $D_1$ is empty, then $D$ is entirely contained in $U_2$, so $G = \langle D \rangle \leq U_2$, hence $G = U_2$.

Now suppose that $D_1$ is nonempty.

Note that if $u \in U_2$ and $d \in D_1$, then $d^u$ cannot be in $U_2$. If it were, then $d = (d^u)^{u^{-1}}$ would also be in $U_2$. But this is impossible because $U_2$ and $D_1$ are disjoint. Therefore, conjugation by $u \in U_2$ induces a permutation of $D_1$; in other words, it permutes the generators of $\langle D_1 \rangle$, so it normalizes $\langle D_1 \rangle$. This means that $U_2 \leq N_G(\langle D_1 \rangle)$. For brevity, we write $N = N_G(\langle D_1 \rangle)$.

Since $D_2 \cup D_{12} \subseteq U_2$, the preceding paragraph implies that $D_2 \cup D_{12} \subseteq N$. Also, certainly $\langle D_1 \rangle$ normalizes itself, so $D_1 \subseteq \langle D_1 \rangle \leq N$. Therefore, $D = D_1 \cup D_2 \cup D_{12} \subseteq N$. This means that $G = \langle D \rangle \leq N$, so $G = N$.

But this means that $\langle D_1 \rangle \lhd G$. So $\langle D_1 \rangle$ is a normal subgroup of $G$ which contains at least one element of $D$. But the elements of $D$ are conjugate, so any normal subgroup containing an element of $D$ must contain all of $D$, and therefore all of $\langle D \rangle = G$. This means that $G \leq \langle D_1 \rangle \leq U_1$, and so $U_1 = G$.