# Describe Cosets of subgroup $H= \lbrace 2^n : n \in \mathbb{Z} \rbrace$ of $\mathbb{R}^*$

"A Book of Abstract Algebra" presents this exercise:

Describe the cosets of the sub-group:

Subgroup $$H= \lbrace 2^n : n \in \mathbb{Z} \rbrace$$ of $$\mathbb{R}^*$$.

EDIT #2

Thanks to Matt Samuel's comment for explaining $$R^*$$:

Usually it denotes the group of nonzero real numbers under multiplication.

END EDIT

I wrote a few values of $$H$$:

$$H = \lbrace 2^0, 2^1, 2^2\rbrace = \lbrace 1, 2, 4 \rbrace$$

But, I'm unsure of how to reason about the cosets of $$H$$ given the infinitely sized, i.e. $$R^*$$, $$a$$ in the coset $$aH$$.

Please point me in the right direction of this problem.

• Your assumption/guess is wrong, and you can't possibly solve the problem without knowing what $\mathbb{R}^{\ast}$ and its group operation are. Maybe you should look up the definition before trying to solve the problem (certainly you shouldn't guess). Commented Jul 20, 2015 at 2:23
• Usually it denotes the group of nonzero real numbers under multiplication. Commented Jul 20, 2015 at 2:27
• $\mathbb{R}^*$ is the multiplicative group of nonzero real numbers, i.e. $(\mathbb{R}\setminus\{0\},\cdot)$. Here, $$H = \left\{\ldots, \frac{1}{4}, \frac{1}{2}, 1, 2, 4,\ldots\right\}.$$ Commented Jul 20, 2015 at 2:27

So the operation on $\mathbb R^\times$ is multiplication. Any coset of $H$ will be $$aH=\{a2^n:n\in\mathbb Z, a\neq 0\}$$

Note that if $a\in H$ then $aH=H$

EDIT: a few notes since it looks like this is all new to you. $\mathbb R^\times$ is the group of real numbers closed under multiplication. Since $0$ has no multiplicative inverse (that is, nothing times zero gives one), $0$ can't be in the group.

Also, there is not "the" coset of $H$, there are many different cosets (in particular, infinite) of $H$. One example here would be $3H=\{3\cdot2^n:n\in\mathbb Z\}$.

$\mathbb{R}^{\ast}$ is usually the set of non-zero real numbers under multiplication, of which $H$ is a subgroup. One way to find cosets is begin with the definition: $aH = bH$ if and only if $b^{-1}a \in H$ which in your case means that $$\exists n\in \mathbb{N} \text{ such that } a = b2^n$$ So, for instance, $$\ldots = (3/4)H = (3/2)H = 3H = 6H = 12H = \ldots$$ Perhaps the book is looking for something better, but I hope this helps.

Let $$C_1 = \{2^n: n\in \mathbb{Z}\}$$. Then the set of all the cosets of the subgroup $$H$$ of $$\mathbb{R}^*$$ is

$$C = \{C_1\} \cup \{C_1r: r\in \mathbb{R^*} \wedge r\not\in C_1\}$$.

The cosets will be

$H$ and $mH$ where $m\neq 2^p;p\in \mathbb Z$

• Could you please show how you came up with $m \in 2\mathbb{Z} + 1$? Commented Jul 20, 2015 at 2:29
• I think you probably misunderstood, @learnmore. We're talking about real numbers here, not just integers. Commented Jul 20, 2015 at 2:31
• @KevinMeredith;i edited it Commented Jul 20, 2015 at 2:38