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UPSC 2017 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Cyclic groups · Algebra · asked 8× in 13 yrs · Read the full method →

Question

Show that the groups $\mathbb Z_5\times\mathbb Z_7$ and $\mathbb Z_{35}$ are isomorphic.

Technique

Show the product group is cyclic by finding an element of order $\operatorname{lcm}(5,7)=35=|G|$ (the generator $(1,1)$); equivalently the CRT isomorphism $x\mapsto(x\bmod5,x\bmod7)$ with trivial kernel.

Solution

Both groups have order $35$. We show $\mathbb Z_5\times\mathbb Z_7$ is cyclic (then any cyclic group of order $35$ is isomorphic to $\mathbb Z_{35}$). We give two complete arguments.

Method 1 — Exhibit a generator (order of $(1,1)$)

Consider the element $(1,1)\in\mathbb Z_5\times\mathbb Z_7$ (the group operation is coordinatewise addition mod $5$ and mod $7$). For $k\ge1$,

This equals the identity $(0,0)$ iff $k\equiv0\pmod5$ and $k\equiv0\pmod7$, i.e. iff $5\mid k$ and $7\mid k$. The smallest such positive $k$ is

Hence $(1,1)$ has order $35$. Since $|\mathbb Z_5\times\mathbb Z_7|=35$, the cyclic subgroup $\langle(1,1)\rangle$ already has $35$ elements and equals the whole group. So $\mathbb Z_5\times\mathbb Z_7$ is cyclic of order $35$, hence

Method 2 — Explicit isomorphism (CRT)

Define $\psi:\mathbb Z_{35}\to\mathbb Z_5\times\mathbb Z_7$ by

• Well defined & homomorphism: if $x\equiv x'\pmod{35}$ then $x\equiv x'$ mod $5$ and mod $7$, and reduction is additive: $\psi(x+y)=\psi(x)+\psi(y)$.
• Injective: if $\psi(x)=(0,0)$ then $5\mid x$ and $7\mid x$; since $\gcd(5,7)=1$, $35\mid x$, so $x\equiv0\pmod{35}$. Thus $\ker\psi=\{0\}$.
• Surjective / bijective: an injective map between finite sets of equal size $|\mathbb Z_{35}|=35=|\mathbb Z_5\times\mathbb Z_7|$ is a bijection.

Hence $\psi$ is a group isomorphism. (This $\psi$ is exactly the Chinese Remainder Theorem isomorphism for coprime moduli.)

Answer