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I am trying to prove that if $G = \mathbb{R}$ then $H = \{\log a \mid a \in \mathbb{Q}, a > 0\}$ is a subgroup.

  1. The identity of $G$ is $0$ and $0 \in H$.
  2. If $a,b \in \mathbb{Q}$ and $a>0$ and $b>0$ then $\log ab = \log a + \log b$.
  3. If $x \in H$ then $x^{-1} \in H$.

I am having trouble convincing myself of the last one. That would mean that if $\log a = y$ then $a = 10^y$ for some $a \in \mathbb{Q}$ and $a >0$.

Am I going about this the right way?

Michael Hardy
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spitfiredd
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2 Answers2

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Let $h \in H\leq G$. Then there exists an $a\gt 0, a\in \mathbb Q$ such that $h = \log a$.

Since the operation of the group is addition, we to determine whether there is an additive inverse $h^{-1} \in H.\;$ Since $\,a > 0,\; a\in \mathbb Q,\,$ so is $\frac 1a \in \mathbb Q, \;\frac 1a >0.\;$ Then $$h'= \log\left(\frac 1a\right)\in H.$$

I'll leave it to you to show that given $$h = \log a \in H \implies h' = h^{-1} = \log\left(\frac 1a\right)\in H$$

amWhy
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The operation you are considering on $\mathbb{R}$ is addition, so you don't want to look at $x^{-1}$, but at $-x$.

There's a slicker way to prove the claim. The map $$ f\colon \mathbb{Q}_{>0}\to\mathbb{R},\quad x\mapsto \log x $$ is a group homomorphism, when $\mathbb{Q}_{>0}$ (the positive rationals) is endowed with multiplication and $\mathbb{R}$ with addition, since $$ f(xy)=\log(xy)=\log x+\log y=f(x)+f(y) $$ and so its image is a subgroup.

egreg
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