Introduction

The group $(\mathbb{R}, +)$ is an abelian, so every subgroup $H \triangleleft \mathbb{R}$ and the quotient group $\mathbb{R}/H$ is well-defined. We are interested in the number of left cosets in these quotient groups, or the index $[\mathbb{R} : H]$. In particular, we want to determine when the index is finite, countable, or uncountable. Does the quotient of a countable subgroup with $\mathbb{R}$ result in a uncountable index? Likewise, does the quotient of an uncountable subgroup with $\mathbb{R}$ result in a countable index? In terms of cardinality, is the number of subgroups with a countable number of left cosets equal to the number of subgroups with an uncountable number of left cosets?

Subgroups with Finite Index

Observe the only subgroup of $\mathbb{R}$ with finite index is $\mathbb{R}$ itself. In particular, $[\mathbb{R} : \mathbb{R}] = 1$.

PROOF

Suppose for the sake of a contradiction that $H$ has finite index, $H \neq \mathbb{R}$, and let $r \in \mathbb{R}$ and $r \notin H$. We will construct inifintely many distinct left cosets of H to obtain the contradiction. We proceed by induction on n and consider cosets of the form $nr + H$. When n = 0, observe $0 + H = H \neq r + H$, since $r \notin H$. Now, if n=k, then $kr + H \neq (k+1)r + H$, since otherwise $(k+1)r \in kr + H$ by properties of cosets. This means $(k+1)r = kr + h$ for some $h \in H$, so $r=h$ and $r \in H$, a contradiction. Thus, $kr + H \neq (k+1)r + H$ and the proof is complete. $ \qquad \blacksquare $

Subgroups with Uncountable Index

We now give an example of a subgroup with uncountable index and show in this particular case, the quotient of a countable subgroup with $\mathbb{R}$ results in an uncountable index. Let $H = \mathbb{Z}$ and consider $\mathbb{R}/\mathbb{Z}$. It is known that $\mathbb{R}/\mathbb{Z} \cong S^1$, the circle group, which contains an uncountable number of elements. Thus, in this example the question is answered affirmatively, and in the following theorem we answer the question more generally.

PROOF

In order to obtain a contradiction, suppose that $H$ is countable and $[\mathbb{R} : H]$ is countable. Since the cosets of $H$ partition $R$, using cardinal arithmetic we can write $|\mathbb{R}| = |H| \times [\mathbb{R} : H]$. Since $|H|$ and $[\mathbb{R} : H]$ are both countable, their product is countable, so this equation states that $\mathbb{R}$ is countable, a contradiction. $\qquad \blacksquare$

Subgroups with Countable Index

Constructing an example of a subgroup with countable index is more difficult than the uncountable case and requires the Axiom of Choice. Since $\mathbb{R}$ is an extension field of $\mathbb{Q}$, we can view $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Thus, we can construct a Hamel basis $\mathcal{B}$, which will be a linearly independent, uncountable set of real numbers such that every real number can be written as a unique, finite linear combination of these basis elements. To construct the subgroup $H$, we select one element $e_0 \in \mathcal{B}$ and define $H = span_\mathbb{Q}(\mathcal{B} - \{ e_0 \})$. It is easy to see that $H$ is a subgroup of ($\mathbb{R}, +)$, since it is a subspace of $\mathbb{R}$ over $\mathbb{Q}$. Thus $[\mathbb{R} : H]$ will be equal to the cardinality of the $span(e_0)$ over $\mathbb{Q}$, which in countably infinite.

Given this example, one might expect that all uncountable subgroups have countable index. However, this is not the case as the following example shows. We again view $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and use a Hamel basis $\mathcal{B}$. Because $\mathcal{B}$ has cardinality $\mathfrak{c}$, we can split it into two disjoint subsets, $X$ and $Y$, such that both $X$ and $Y$ also have cardinality $\mathfrak{c}$.$$\mathcal{B} = X \cup Y \quad \text{where} \quad X \cap Y = \emptyset \quad \text{and} \quad |X| = |Y| = \mathfrak{c}$$ Define $H$ as the rational span of the first half of the basis. $$H = \text{span}_{\mathbb{Q}}(X)$$Because $X$ contains uncountably many linearly independent elements, $H$ is uncountably large. Therefore, $H$ is an uncountable subgroup of $\mathbb{R}$. The index of $H$ is the cardinality of the quotient group $\mathbb{R}/H$. By the definition of our partition, the quotient group $\mathbb{R}/H$ is isomorphic to the rational span of the remaining basis elements, $Y$. $$\mathbb{R}/H \cong \text{span}_{\mathbb{Q}}(Y)$$Since $Y$ also has cardinality $\mathfrak{c}$, its span has cardinality $\mathfrak{c}$. Therefore, the quotient group has $2^{\aleph_0}$ elements, meaning the index $[\mathbb{R} : H]$ is uncountable. Thus, the two examples prove the following statement.

Comparing the Number of Subgroups with Countable and Uncountable Index

In terms of cardinality, is the number of subgroups with countable index in $(\mathbb{R}, +)$ equal to the number of subgroups with uncountable index? We suspect that the answer is yes but will not prove it here. Maybe later? For now, we leave it as a conjecture.

Conclusion

We've shown that the only normal subgroup of $(\mathbb{R}, +)$ with finite index is $\mathbb{R}$ itself. We also showed that countable subgroups have uncountable index. The case was less simple for uncountable subgroups and required the Axiom of Choice to construct one. We provided examples that show the index of an uncountable subgroup can either be countable or uncountable. We left unanswered the question of whether the cardinality of the number of subgroups with countable index in $(\mathbb{R}, +)$ is equal to that of the subgroups with uncountable index.

References

Enderton, H. B. (1977). Elements of set theory. Academic Press.

[2] Dummitt, D.S. and Foote, R.M. (2004) Abstract Algebra. 3rd Edition, John Wiley & Sons, Inc.

[3] Gallian, J. (2021). Contemporary Abstract Algebra (10th ed.). Chapman and Hall/CRC