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UPSC 2018 Maths Optional Paper 2 Q3a — Step-by-Step Solution

20 marks · Section A

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

Question

Find all the proper subgroups of the multiplicative group of the field $(\mathbb Z_{13},+_{13},\times_{13})$, where $+_{13}$ and $\times_{13}$ represent addition modulo 13 and multiplication modulo 13 respectively.

Technique

Finite-field multiplicative group is cyclic; subgroup lattice of $\mathbb Z_{12}$ — one subgroup per divisor $d\mid12$, generated by $2^{12/d}$.

Solution

The multiplicative group is $G=\mathbb Z_{13}^\times=\{1,2,3,\ldots,12\}$ under multiplication mod $13$ (since $13$ is prime, every nonzero residue is invertible). Thus $|G|=12$.

Step 1 — $G$ is cyclic of order $12$; find a generator

The multiplicative group of any finite field is cyclic, so $G$ is cyclic of order $12$. We show $2$ is a generator by computing its powers mod $13$:

The powers $\{2^1,\ldots,2^{12}\}=\{2,4,8,3,6,12,11,9,5,10,7,1\}$ run through all $12$ elements, so $\operatorname{ord}(2)=12$ and

Step 2 — Subgroups of a cyclic group: one per divisor

A cyclic group of order $12$ has exactly one subgroup of order $d$ for each divisor $d\mid 12$, namely $\langle g^{12/d}\rangle$ where $g=2$. The divisors of $12$ are

The subgroup of order $12$ is $G$ itself (improper). The proper subgroups correspond to the proper divisors $d\in\{1,2,3,4,6\}$ — five proper subgroups.

Step 3 — List each proper subgroup explicitly

Using $H_d=\langle 2^{12/d}\rangle$ of order $d$:

For example $H_4=\langle 8\rangle$: $8^1=8,\ 8^2=64\equiv12,\ 8^3\equiv96\equiv5,\ 8^4\equiv40\equiv1$, giving $\{1,8,12,5\}$. And $H_6=\langle4\rangle$: $4,16\equiv3,12,48\equiv9,36\equiv10,40\equiv1$, i.e. $\{4,3,12,9,10,1\}$ — the quadratic residues mod $13$.

Answer