← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

Isomorphism theorems (First, Second, Third) · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Show that the quotient group of $(\mathbb R,+)$ modulo $\mathbb Z$ is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here $\mathbb R$ is the set of real numbers and $\mathbb Z$ is the set of integers.

Technique

First Isomorphism Theorem — exhibit a surjective homomorphism $\mathbb R\to S^1$, $x\mapsto e^{2\pi i x}$, and compute its kernel $=\mathbb Z$.

Solution

Let $S^1=\{z\in\mathbb C:|z|=1\}$ be the unit circle, a group under multiplication. We must show

Step 1 — $\mathbb Z$ is a normal subgroup, so $\mathbb R/\mathbb Z$ is a group

$(\mathbb R,+)$ is abelian, and $\mathbb Z\le\mathbb R$ is a subgroup. Every subgroup of an abelian group is normal, so the quotient group $\mathbb R/\mathbb Z$ is well-defined, with cosets $x+\mathbb Z$ and operation $(x+\mathbb Z)+(y+\mathbb Z)=(x+y)+\mathbb Z$.

Step 2 — Define a homomorphism $\varphi:\mathbb R\to S^1$

Define

$\varphi$ maps into $S^1$: $|e^{2\pi i x}|=1$ for all real $x$. ✓ $\varphi$ is a homomorphism from $(\mathbb R,+)$ to $(S^1,\times)$:

Step 3 — $\varphi$ is surjective

Any point of $S^1$ has the form $e^{i\theta}=\cos\theta+i\sin\theta$ for some $\theta\in[0,2\pi)$. Take $x=\dfrac{\theta}{2\pi}\in[0,1)$; then $\varphi(x)=e^{2\pi i\cdot\theta/(2\pi)}=e^{i\theta}$. Hence every element of $S^1$ is attained: $\varphi$ is onto.

Step 4 — Compute the kernel

Now $e^{2\pi i x}=1\iff 2\pi x=2\pi n$ for some integer $n\iff x=n\in\mathbb Z$. Therefore

Step 5 — First Isomorphism Theorem

$\varphi:\mathbb R\to S^1$ is a surjective homomorphism with kernel $\mathbb Z$. By the First Isomorphism Theorem,

$\blacksquare$

The induced isomorphism is $\bar\varphi:\mathbb R/\mathbb Z\to S^1,\ \bar\varphi(x+\mathbb Z)=e^{2\pi i x}$, which is well-defined (independent of coset representative: if $x'=x+n$ then $e^{2\pi i x'}=e^{2\pi i x}e^{2\pi i n}=e^{2\pi i x}$), bijective, and multiplicative.

Verification

• Well-definedness on cosets: two representatives $x,\,x+n$ ($n\in\mathbb Z$) give the same image since $e^{2\pi i n}=1$. ✓ This is exactly the kernel condition, confirming the map descends to the quotient.
• Group-axiom spot check: identity $0+\mathbb Z\mapsto e^0=1$ (identity of $S^1$); inverse $(-x)+\mathbb Z\mapsto e^{-2\pi i x}=\overline{e^{2\pi i x}}=(e^{2\pi i x})^{-1}$. ✓
• Element-order match (numerical): the coset $\tfrac14+\mathbb Z$ has order $4$ in $\mathbb R/\mathbb Z$ ($4\cdot\tfrac14=1\in\mathbb Z$); its image $e^{2\pi i/4}=i$ has order $4$ in $S^1$ ($i^4=1$). Orders are preserved, as required of an isomorphism. ✓