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Consider $G$ a topological group and $S$ a connected subset of this group. I want to show that $<S>$, the group generated by $S$ is also connected. For each $n$, I can construct $2^n$ continuous maps :

$$\phi_{ni}:\underbrace{C\times \ldots \times C}_{n} \to G$$ One of those map would for example look like $\phi_{42}(c_1,c_2,c_3,c_4)=c_1^{-1}c_2c_3c_4$.

Then I can write $$<S> = \bigcup\limits_{n=1}^{\infty}\bigcup\limits_{i=1}^{2^n}\phi_{ni}(\underbrace{C\times \ldots \times C}_n)$$ where the $\phi_{ni}(\underbrace{C\times \ldots \times C}_n)$ are connected since the $\phi$'s are connected. But then why is $<S>$ connected?

roi_saumon
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    This is not true in general, take $\mathbb{Z}$ as a topological group and take the subgroup generated by $S={2}$, then $S$ is connected but $=2\mathbb{Z}$ is not.. – M. Van Jul 14 '20 at 14:44
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    As a discrete group, $\mathbb{Z}$ is generated by a singleton set ${1}$, which is connected. But $\mathbb{Z}$ is not connected in the discrete topology. – halrankard Jul 14 '20 at 14:44
  • I guess you need some extra assumptions on $S$. – M. Van Jul 14 '20 at 15:03

1 Answers1

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If $S$ contains the identity element, then this is true. To see this, first observe that in this case $S\cup S^{-1}$ is connected, where $S^{-1}=\{s^{-1}|s\in S\}$. This is because $S^{-1}$ is connected (as the image of $S$ under the map $g\mapsto g^{-1}$ and the intersection $S\cap S^{-1}$ is nonempty (it contains the identity. Now, set $R=S\cup S^{-1}$. Then $\bigcup_{n\in \mathbb{N}} R^n$, is connected where $R^n=\{s_1\cdots s_n | s_1,...,s_n\in R\}$, since each $R^n$ is connected (as the image of $R\times\cdots R$ under the map $(g_1,...,g_n)\mapsto g_1\cdots g_n$), and their intersection is nonempty (the identity element is in it). Thus $G$ is connected, since $G=\mathrm{Cl}(\bigcup_{n\in \mathbb{N}} R^n)$, and the closure of a connected set is connected.

This is also true if $S$ generates $G$ as a semigroup. That is, if $G$ is the only closed sub-semigoup of $G$ containing $S$, and $S$ is connected, then $G$ is connected as well.

Cronus
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