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I already know that when G is a connected solvable Hausdorff group there exists a sequence

$$ G = G_0 \rhd G_1 \rhd ... \rhd G_R = (e)$$

where the $G_i$ are closed connected and $G_{i-1}/G_i$ is abelian. Furthermore I'm aware of how to prove that $G_{i-1}/G_i \simeq \mathbb{T}^{a_i} \times \mathbb{R}^{b_i } $ since it is abelian.

I can't tell how to show that $G_i$ can be chosen so that $G_{i-1}/G_i \simeq \mathbb{T}$ or $G_{i-1}/G_i \simeq \mathbb{R}$. I was told to take the inverse image in $G_{i-1}$ of the sequence

$$ (0) \subset \mathbb{T} \subset ... \subset \mathbb{T}^{a_i} \subset \mathbb{T}^{a_i} \times \mathbb{R} \subset ... \subset \mathbb{T}^{a_i} \times \mathbb{R}^{b_i} $$

but don't really know how to make sense of that. Any help would be greatly appreciated.

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1 Answer 1

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To be precise, given $\mathbb{T}^a\times\mathbb{R}^b$ we can define

$$H_j=(0)^{a-j}\times \mathbb{T}^j\times (0)^b$$ $$K_j=\mathbb{T}^a\times (0)^{b-j}\times\mathbb{R}^j$$

In this way both $H_j$ and $K_j$ are subgroups of $\mathbb{T}^a\times\mathbb{R}^b$ and the following sequence

$$(0)\subseteq H_1\subseteq H_2\subseteq\cdots\subseteq H_a\subseteq K_1\subseteq\cdots\subseteq K_b=\mathbb{T}^a\times\mathbb{R}^b$$

has the property that consecutive quotients are either $\mathbb{T}$ or $\mathbb{R}$.

Now we have two consecutive subgroups $G_{i}\subset G_{i-1}$ and an isomorphism $F:G_{i-1}/G_i\to\mathbb{T}^a\times\mathbb{R}^b$. Consider the projection $\pi:G_{i-1}\to G_{i-1}/G_i$. Then we have the composed map $\tau:G_{i-1}\to\mathbb{T}^a\times\mathbb{R}^b$, $\tau=F\circ\pi$ which is surjective, with $G_i$ as its kernel. And this map induces the following refined sequence:

$$G_{i}=\tau^{-1}(0)\subseteq \tau^{-1}(H_1)\subseteq \tau^{-1}(H_2)\subseteq\cdots\subseteq \tau^{-1}(H_a)\subseteq $$ $$\subseteq\tau^{-1}(K_1)\subseteq\cdots\subseteq \tau^{-1}(K_b)=G_{i-1}$$

We've just applied $\tau^{-1}$ to every subgroup in the previous sequence. It is the consequence of the third isomorphism theorem that consecutive quotients are either $\mathbb{T}$ or $\mathbb{R}$.

And so you add those additional groups between every $G_i$ and $G_{i-1}$. So it is not that you can choose $G_i$ in such way. You can refine the sequence. In fact, it is not always possible if $R$ (the initial length of the sequence) is fixed. In other words: you can choose $G_i$ in such way, but you have no control over $R$.

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  • $\begingroup$ Thank you this helps a lot. Some points I would like to clarify: Can we assume that our refined sequence is still closed connected and normal due to $\pi$ being an open, continuous homomorphism? And would this be the correct way to apply the third isomorphism theorem: $\tau^{-1}(K_{j-1})/ \tau^{-1}(K_j) \simeq (\tau^{-1}(K_{j-1})/G_i)/( \tau^{-1}(K_j)/G_i) \simeq K_{j-1}/ K_j \simeq \mathbb{R}$ where we chose $K_j$ instead of $H_j$ w.l.o.g. $\endgroup$
    – diesmond
    Commented Aug 10, 2023 at 7:53
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    $\begingroup$ @diesmond these are closed, because $H_j$, $K_j$ are closed and $\tau$ is continuous. These are normal, because preimage (via homomorphism) of normal subgroup is normal. These are connected by the fact that given $A\subseteq B$ we have that $B$ is connected if and only if $A$ and $B/A$ are, plus induction. Finally, yes: this is the correct way to apply the third isomorphism theorem. $\endgroup$
    – freakish
    Commented Aug 10, 2023 at 9:22

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