Is it right that $\cup_{g\in G}gHg^{-1}\subsetneq G$, for every $H$, proper subgroup of infinite group $G$?

Question

Let $G$ be a group and $H$, a proper subgroup of $G$ with $[G:H]<\infty$. Is it right that $\bigcup\limits_{g\in G}gHg^{-1}$ is a proper subset of $G $? If $G$ is a finite group we can prove this claim, but is this right for infinite groups?

Answer 1

It is false for some infinite groups. (Edit: The "false" here refers to the question asked in the title, not in the body where there is the additional condition of $H$ having finite index in $G$.) The simplest counterexample is $G = {\rm GL}_2(\mathbf C)$ and $H$ is the subgroup of upper-triangular matrices $\begin{pmatrix}a&b\\0&c\end{pmatrix}$ where $a, c \in \mathbf C^\times$.

Every $A \in {\rm GL}_2(\mathbf C)$ has an eigenvector in $\mathbf C^2$, say $v$ with eigenvalue $\lambda$: $Av = \lambda v$ and $v \not= \binom{0}{0}$. Let $w$ be a vector in $\mathbf C^2$ that is outside the line $\mathbf C v$. We can write $Aw = zv + z'w$ for $z$ and $z'$ in $\mathbf C$. (The matrix $A$ might not have an eigenvector linearly independent of $v$, i.e., not all $2 \times 2$ complex matrices are diagonalizable, so we need not be able to pick $w$ as an eigenvector of $A$.) The matrix representation of $A$ with respect to the basis $\{v,w\}$ of $\mathbf C^2$ is $\begin{pmatrix}\lambda &z\\0&z'\end{pmatrix}$, so $A$ is conjugate by an invertible matrix in $G$ to a $2 \times 2$ matrix in $H$. That proves $G = \bigcup_{g \in G} gHg^{-1}$.

Answer 2

Yes it is correct. Consider the action of $G$ by left multiplication on the finite set $X$ of left cosets of $H$ in $G$. This action is transitive. The action defines a homomorphism $\phi:G \to {\rm Sym}(X)$.

Then image $P$ of $\phi$ is a transitive group group on the finite set $X$, and there must exist an element $a \in P$ without fixed points. (This follows from the result that the average number of fixed points in a transitive action on a finite set is $1$, and the identity fixes more than one point - since $H$ is a proper subgroup we have $|X| > 1$).

Now, if $b \in gHg^{-1}$ for some $g \in G$, then $b(gH) = gH$, so $\phi(b)$ fixes $gH$. Hence an element $b \in G$ with $\phi(b) = a$ does not lie in $\cup_{g \in G} gHg^{-1}$.