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UPSC 2013 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Cyclic groups · Algebra · asked 8× in 14 yrs · Read the full method →

Question

Give an example of an infinite group in which every element has finite order.

Technique

Standard counterexample to “infinite group ⇒\Rightarrow has element of infinite order” — both Q/Z\mathbb{Q}/\mathbb{Z} and roots of unity work.

Solution

Example. G=Q/ZG=\mathbb{Q}/\mathbb{Z} — the additive group of rationals modulo integers.

Step 1 — GG is an infinite group

Elements of GG are cosets r+Zr+\mathbb{Z} with r∈Qr\in\mathbb{Q}. Two cosets r+Zr+\mathbb{Z} and s+Zs+\mathbb{Z} are equal iff r−s∈Zr-s\in\mathbb{Z}. So a canonical representative is r∈[0,1)∩Qr\in[0,1)\cap\mathbb{Q}.

The set [0,1)∩Q[0,1)\cap\mathbb{Q} is infinite (contains 1/n1/n for each n≥2n\ge 2), so GG is infinite.

Step 2 — Every element has finite order

Let g=r+Z∈Gg=r+\mathbb{Z}\in G, written as r=p/qr=p/q in lowest terms with p,q∈Z,  q≥1p,q\in\mathbb{Z},\;q\ge 1. Then

q⋅g=qr+Z=p+Z=0+Z=0G,q\cdot g=qr+\mathbb{Z}=p+\mathbb{Z}=0+\mathbb{Z}=0_G,

since p∈Zp\in\mathbb{Z}. So the order of gg divides qq — finite (specifically, equal to q/gcd⁡(p,q)=qq/\gcd(p,q)=q for p/qp/q in lowest terms with gcd⁡(p,q)=1\gcd(p,q)=1).

Answer

  G=Q/Z is an infinite group with every element of finite order.  \boxed{\;G=\mathbb{Q}/\mathbb{Z}\text{ is an infinite group with every element of finite order.}\;}
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