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UPSC 2016 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

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

Question

Let $p$ be a prime number and $\mathbb Z_p$ denote the additive group of integers modulo $p$. Show that every non-zero element of $\mathbb Z_p$ generates $\mathbb Z_p$.

Technique

Lagrange’s theorem forces the subgroup order to be $1$ or $p$; nonzero rules out $1$. (Backup self-contained proof via $\gcd(a,p)=1$ and Euclid’s lemma.)

Solution

$\mathbb Z_p=\{0,1,\ldots,p-1\}$ under addition mod $p$ is a finite group of order $p$. We show that any nonzero $a$ has additive order $p$, hence the cyclic subgroup it generates is all of $\mathbb Z_p$. Two clean routes are given; both use that $p$ is prime.

Step 1 — Setup: order of an element divides the group order

Let $a\in\mathbb Z_p$, $a\ne0$ (so $1\le a\le p-1$). The cyclic subgroup generated by $a$ is

where $ka$ means $a$ added to itself $k$ times. By Lagrange’s theorem, $|\langle a\rangle|$ divides $|\mathbb Z_p|=p$. Since $p$ is prime, its only positive divisors are $1$ and $p$. So $|\langle a\rangle|\in\{1,p\}$.

Step 2 — Rule out the trivial case

$|\langle a\rangle|=1$ would mean $\langle a\rangle=\{0\}$, i.e. $a=0$. But $a\ne0$. Hence

and since $\langle a\rangle\subseteq\mathbb Z_p$ with equal (finite) cardinality, $\langle a\rangle=\mathbb Z_p$. Therefore $a$ generates $\mathbb Z_p$.

Answer