← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Subgroups; Subgroup Criterion · Algebra · Read the full method →

Question

Show that the additive group $\mathbb Q$ of rational numbers has infinitely many subgroups.

Technique

Exhibit explicit infinite family $H_n=(1/n)\mathbb Z$; show distinctness via subgroup relation $H_n\subseteq H_m\Leftrightarrow n\mid m$.

Solution

Strategy. Construct infinitely many distinct subgroups of $(\mathbb Q,+)$.

Construction 1 — Cyclic subgroups generated by $1/n$

For each positive integer $n$, define $H_n=\langle 1/n\rangle=\{k/n:k\in\mathbb Z\}=\dfrac{1}{n}\mathbb Z$.

$H_n$ is a subgroup:

• $0=0/n\in H_n$.
• Closed under +: $(k_1/n)+(k_2/n)=(k_1+k_2)/n\in H_n$.
• Closed under inverses: $-(k/n)=(-k)/n\in H_n$.

Step 1 — Show $H_n$‘s are distinct

$H_n\ne H_m$ when $n\ne m$ (e.g., when one isn’t divisible by the other).

Actually $H_n\subseteq H_m$ iff $1/n\in H_m=\dfrac{1}{m}\mathbb Z$, i.e., $1/n=k/m$ for some integer $k$, i.e., $m=nk$, i.e., $n\mid m$.

So $H_n\subseteq H_m\Leftrightarrow n\mid m$.

Distinct $n$‘s give distinct chains, but two $H_n$‘s can be equal if and only if $n=m$ (since $H_n\subseteq H_m$ and $H_m\subseteq H_n$ requires $n\mid m$ and $m\mid n$, hence $n=m$).

So $\{H_n:n\in\mathbb Z^+\}$ is an infinite family of distinct subgroups of $\mathbb Q$.

Answer