← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Cosets and Lagrange's theorem · Algebra · asked 4× in 13 yrs · Read the full method →

Question

Let $G$ be a finite group, $H$ and $K$ subgroups of $G$ such that $K\subset H$. Show that $(G:K)=(G:H)(H:K)$.

Technique

Lagrange’s theorem (index = order quotient) for the one-line finite proof; coset-partition bijection for the general statement.

Solution

Here $K\subseteq H\subseteq G$ are subgroups, so $K$ is a subgroup of $H$ as well as of $G$. The index $(G:K)$ denotes the number of (left) cosets of $K$ in $G$.

Step 1 — Lagrange’s theorem gives the orders

Since $G$ is finite, Lagrange’s theorem applies to every subgroup. For a subgroup $A\le B$ of a finite group, $(B:A)=|B|/|A|$. Hence

($H$ is itself a finite group and $K\le H$, so $(H:K)=|H|/|K|$ is legitimate.)

Step 2 — Combine

Answer