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For any $k \in \mathbb{R^*}, \langle k\rangle$ is a normal subgroup.

Consider the case $k \neq 1, k>0$. Then, via the surjective homomorphism $$\varphi: \mathbb{R^*} \to \mathbb{T}\times\{\pm1\}, \quad x\mapsto (e^{2\pi i\cdot \log_{k}(|x|)}, \: \textrm{sign}(x))$$

$$\implies \mathbb{R^*}/\langle k\rangle \simeq \mathbb{T} \times\{\pm1\}.$$

by the First Isomorphism Theorem (where $\mathbb{T}$ is the circle of unit radius in the complex plane).

But for any $M, N \trianglelefteq G$, $$\langle M, N\rangle = MN \trianglelefteq G$$

$$\implies \langle k_1,...,k_n\rangle \trianglelefteq \mathbb{R^*}. $$

There turns out to be a few different cases for $n=1$, so suppose that $k_1, ... k_n \in \mathbb{R^*},$ with $k_i \neq 1, k_i >0$.

To which group is $\mathbb{R^*}/\langle k_1,...,k_n\rangle$ isomorphic? I can't spot a friendly homomorphism for this one so easily.

legionwhale
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  • Is $\mathbf{R}^*$ the multiplicative group of the positive real numbers? If so, $\log_k$ is not defined if $k = 1$; so, we don’t have a surjective homomorphism (or even a homomorphism at all) in that case. – shoteyes Jun 28 '21 at 02:06
  • Note that $\mathbb R^*\cong\mathbb R_{>0}\times{\pm1}\cong\mathbb R\times C_2$. – Kenta S Jun 28 '21 at 02:16
  • @shoteyes You are right. If $k = 1$, then the quotient is isomorphic to $\mathbb{R^*}$. Thanks for pointing that out. – legionwhale Jun 28 '21 at 12:37
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    If $\mathbf{R}^$ is supposed to be the nonzero reals as @KentaS says, then the $\log_k x$ homomorphism is not defined for negative $x$. If, instead, you mean $\mathbf{R}^$ to be the positive reals, then it’s fine. – shoteyes Jun 28 '21 at 12:58
  • @shoteyes Yikes, that's a big one. And, also $\log_k$ is not defined on the reals for $k < 0$, right? OK, so is the isomorphism to $\mathbb{R} \times {\pm1}$ via $\varphi : x \mapsto (\ln(|x|),$ sign$(x))$ fine then? (I'm not sure why it's $\mathbb{R_{>0}}$). But this seems to make it even more difficult to find the quotient. – legionwhale Jun 28 '21 at 13:04
  • That’ll work to show that $\mathbf{R}^$ is isomorphic to $\mathbf{R} \times {\pm 1}$. Here’s the issue: $$\mathbf{R}^/\langle k\rangle\cong \begin{cases}\mathbf{R}^*, & k = 1 \ \mathbf{R}^{>0}, & k = -1 \ \mathbf{T}\times {\pm 1}, & k > 0, k \neq 1 \ \mathbf{T}, & k < 0, k \neq -1\end{cases}$$ – shoteyes Jun 28 '21 at 13:07
  • @shoteyes I feared that something like this might be the case after I read your first reply, but I hoped it wasn't :( I'll try to find each isomorphism. – legionwhale Jun 28 '21 at 13:16
  • @shoteyes Ok, so using the FIT, first one is identity mapping from $ \mathbb{R^} \to \mathbb{R^}$, next one is mapping to $\mathbb{R^{>0}}$ via modulus, and the last two are similar to the one I gave in the question above, but with a modulus for the $k$, and accounting for sign. So does this mean there isn't going to be a neat closed form for the $\mathbb{R^*} / \langle k_1, ..., k_n \rangle $ quotient? What if I restrict to the case where $k_i > 0, k_i \neq 1$? – legionwhale Jun 28 '21 at 13:22
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    Careful, the surjective homomorphisms you want are $(e^{2\pi i \log_k\lvert x \rvert}, \operatorname{sgn} x)$ and $(\operatorname{sgn} x)e^{\pi i \log_{-k}\lvert x \rvert}$ for the last two cases, respectively. Unfortunately, I cannot think of a clean way to describe the quotient you want. There are too many cases to deal with for me. – shoteyes Jun 28 '21 at 13:27
  • @shoteyes Ah, ok, thank you! The main reason I asked this question is because I was a little bit mystified about constructing a homomorphism that sends even $\langle k_1, k_2 \rangle$ to a kernel (except the trivial case where $\langle k_1, k_2 \rangle = { \pm 1}$) . For $n =1$, $\log$ works handily, but even for $n=2$, I'm not sure what to do. – legionwhale Jun 28 '21 at 13:35

1 Answers1

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Dealing with quotients of $\Bbb{R}_{>0}$ only. Isolating the effect of the factor $\langle -1\rangle\simeq C_2$ should pose no problems.

By uniqueness of positive roots of positive numbers we see that the multiplicative group $\Bbb{R}_{>0}$ is actually isomorphic to the additive group of the vector space $V$ over $\Bbb{Q}$ with uncountable dimension. See Hurkyl's old answer. Alternatively, we see that the multiplicative group $\Bbb{R}_{>0}$ becomes a 1-dimensional vector space over $\Bbb{R}$, if we define the scalar multiplication by $\alpha\star x=x^\alpha$, for all $x>0$ and all $\alpha\in\Bbb{R}$. The singleton set $\{e\}$ is obviously a basis. We can then restrict the scalar multiplication to $\Bbb{Q}$ as we want to only use the structure of a vector space over $\Bbb{Q}$.

The subgroup $G=\langle k_1,\ldots,k_n\rangle$ is a torsion free finitely generated abelian group, hence a free $\Bbb{Z}$-module. So $G$ has a $\Bbb{Z}$-basis $\mathcal{B}$. Assuming the axiom of choice we can extend $\mathcal{B}$ to a $\Bbb{Q}$-basis $\mathcal{C}$ of $V$. Clearly $\mathcal{C}$ is an uncountable set. As $\mathcal{B}$ is finite, $\mathcal{C}\setminus\mathcal{B}$ has the same cardinality as $\mathcal{C}$.

Let $m=|\mathcal{B}|$ be the rank of $G$. It follows that we have an isomorphism $$ V/G\simeq (\Bbb{Q}/\Bbb{Z})^m\times V', $$ where $V'$ is the $\Bbb{Q}$-space spanned by $\mathcal{C}\setminus\mathcal{B}$. Cardinality of a basis dictates the isomorphism type of a vector space, so actually $V'\simeq V$.

We have shown that $$\Bbb{R}_{>0}/\langle k_1,k_2,\ldots,k_n\rangle\simeq (\Bbb{Q}/\Bbb{Z})^m\times\Bbb{R}_{>0}$$ for some positive integer $m\le n$. On the right hand side the group structure is addition in the first factor and multiplication in the second.

The catch is that because AoC was invoked, it is impossible to describe this isomorphism explicitly. We can only deduce that one exists.

It may be mildly counterintuitive that $\Bbb{R}_{>0}$ appears as a factor of its proper quotient group, but infinite dimensional spaces sometimes behave in surprising ways.

If we discard the axiom of choice, then ... I don't know what happens :-)

Jyrki Lahtonen
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