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UPSC 2019 Maths Optional Paper 2 Q2b — Step-by-Step Solution

10 marks · Section A

Normal subgroups; quotient groups · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Write down all quotient groups of the group $\mathbb Z_{12}$.

Technique

Subgroup lattice of a cyclic group (one subgroup per divisor); quotient of cyclic is cyclic with order $|G|/|H|$.

Solution

$\mathbb Z_{12}$ is cyclic (hence abelian), so every subgroup is normal and every subgroup yields a quotient group. We list the subgroups, then the quotients.

Step 1 — Subgroups of $\mathbb Z_{12}$

A cyclic group of order $n$ has exactly one subgroup of order $d$ for each divisor $d\mid n$, namely $\langle n/d\rangle$. The divisors of $12$ are $1,2,3,4,6,12$, so there are six subgroups

| $d=|H_d|$ | generator $12/d$ | subgroup $H_d$ | |---|---|---| | $1$ | $-$ | $\{0\}$ | | $2$ | $6$ | $\{0,6\}$ | | $3$ | $4$ | $\{0,4,8\}$ | | $4$ | $3$ | $\{0,3,6,9\}$ | | $6$ | $2$ | $\{0,2,4,6,8,10\}$ | | $12$ | $1$ | $\mathbb Z_{12}$ |

Step 2 — Form the quotients

For each subgroup $H_d$ of order $d$, the quotient $\mathbb Z_{12}/H_d$ has order $12/d$. A quotient of a cyclic group is cyclic, so $\mathbb Z_{12}/H_d\cong \mathbb Z_{12/d}$.

| Subgroup $H$ | $|H|$ | $|\mathbb Z_{12}/H|$ | Quotient $\mathbb Z_{12}/H$ (up to iso) | |---|---|---|---| | $\{0\}$ | $1$ | $12$ | $\mathbb Z_{12}$ | | $\{0,6\}$ | $2$ | $6$ | $\mathbb Z_{6}$ | | $\{0,4,8\}$ | $3$ | $4$ | $\mathbb Z_{4}$ | | $\{0,3,6,9\}$ | $4$ | $3$ | $\mathbb Z_{3}$ | | $\{0,2,4,6,8,10\}$ | $6$ | $2$ | $\mathbb Z_{2}$ | | $\mathbb Z_{12}$ | $12$ | $1$ | $\{0\}$ (trivial) |

Answer