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Let $G$ be a locally compact group with the group of topological group automorphisms $Aut(G)$ furnished with the compact-open topology. Let $B$ be a subgroup of $Aut(G)$. We call $G$ an [IN$]_B$ if there is a $B$-invariant relatively compact neighbourhood of the identity element of the group $G$.

It is known that (e.g. look at [1, Theorem 2.5]) for an [IN$]_B$ group $G$, the intersection of all $B$-invariant relatively compact neighbourhoods of the identity forms a compact subgroup of $G$, denoted here by $K_B$.

I am looking for a group $G$ with a subgroup $B$ in $Aut(G)$ so that $K_B$ is not a normal subgroup.

Remarks:

  1. One can easily note that the group $B$ cannot include all the inner automorphisms.
  2. By [1, Theorem 0.1], we can conclude that $B$ cannot be relatively compact in $Aut(G)$. Otherwise, $G$ would be [SIN$]_B$ and in this case, $K_B$ is the trivial group of the identity.

[1] Grosser, Siegfried; Moskowitz, Martin; Compactness conditions in topological groups. J. Reine Angew. Math. 246 1971 1–40, DOI: 10.1515/crll.1971.246.1.

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Take $G=\text{GL}_2(\mathbb{Z})\ltimes (\mathbb{R}/\mathbb{Z})^2$ and take $B<G$ to be the $\begin{bmatrix} 1 & * \\ 0 & 1 \end{bmatrix} \simeq \mathbb{Z}$. Then $K_B$ is one of the factors $\mathbb{R}/\mathbb{Z}$.

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  • $\begingroup$ Would $K_B$ be non-normal? It seems central to me. $\endgroup$ – Mahmood Al Jun 15 '17 at 8:16
  • $\begingroup$ it will not commute with the opposite unipotent. $\endgroup$ – Uri Bader Jun 15 '17 at 8:32
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Let $F$ be a non commutative finite group and let $\Bbb{Z}$ be the group of integers. By $\prod_{n \in \Bbb{Z}} F$ we mean the compact group of all $(a_n)_{n \in \Bbb{Z}}$, $a_n \in F$ $\forall n \in \Bbb{Z}$, equipped with the product topology. It is easy to observe that $$ \psi: \Bbb{Z} \rightarrow Aut\left( \prod_{n \in \Bbb{Z}} F\right) $$ where $$ \psi(k)\left((a_n)_n\right)=(a_{n+k})_n, \quad (a_n)_{n \in \Bbb{Z}}\in \prod_{n \in \Bbb{Z}} F. $$ Let $N = \prod_{n \in \Bbb{Z}} F \rtimes_\psi \Bbb{Z}$ be the semi direct product of these two groups.

Let $\bigoplus_{n \in \Bbb{Z}} F$ be the discrete group of the direct product of the groups $F$ index by $\Bbb{Z}$ which is the set of $(b_n)_{n\in \Bbb{Z}}$ where $b_n$ is the identity of $F$ except for finitely many $n \in \Bbb{Z}$.

Note that $\phi$ is a group homomorphism given by $$ \phi: N \rightarrow Aut\big(\bigoplus_{n \in \Bbb{Z}} F\big) $$ where $$ \phi\big( (a_n)_n, k \big) (b_n)_n = (a_n b_{n+k} a_n^{-1}). $$

Let $$ G = \bigoplus_{n \in \Bbb{Z}} F \rtimes_\phi \big( \prod_{n \in \Bbb{Z}} F \rtimes_\psi \Bbb{Z}\big). $$ and let $B$ be the group of inner automorphisms of the elements of subgroup $\{0\} \times \{0\} \times \Bbb{Z}$. Indeed, for each $\beta_\ell \in B$ ($\ell \in \Bbb{Z}$),
$$ \beta_\ell \big( (a_n)_n, (b_n)_n, k\big):= \big(0,0,\ell\big)\big( (a_n)_n, (b_n)_n, k\big)\big(0,0,-\ell\big) = ( (a_{n+\ell}), (b_{n+\ell}), k\big). $$ One can show that $K_B$ (defined in the question) is nothing but the compact group $$ \{0\} \times \prod_{n \in \Bbb{Z}} F \times \{0\}. $$ But $K_B$ cannot be normal because, \begin{eqnarray*} & & \big((a_n)_n, (c_n)_n, k) ( 0 , (b_n)_n, 0\big) \big((a_n)_n, (c_n)_n, k\big)^{-1}\\ &=& \big((a_n)_n, (c_n b_{n+k})_n, k\big) \big( (c_{n-k}^{-1} a_{n-k}^{-1} c_{n-k})_n, (c_{n-k}^{-1}), -k \big)\\ &=& \big( (a_n c_n b_{n+k} c_{n}^{-1} a_{n}^{-1} c_n b_{n+k}^{-1} c_n^{-1})_n, ( c_n b_{n+k}c_n^{-1})_n, 0 \big). \end{eqnarray*} So if $F$ is non-commutative enough (and I have not made any mistakes in my claims), this would be an answer to my own question.

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