GATE 2018 CS Q.19: solution and answer

Official answer: 42Source: Official answer key, GATE 2018 CS (IIT Guwahati)

1. By Lagrange's theorem, the order of any subgroup $H$ of $G$ divides $|G|=84$. A proper subgroup has $|H|<84$.

2. Since $84=2^2\cdot 3\cdot 7$, the divisors of 84 are $1,2,3,4,6,7,12,14,21,28,42,84$. Those below 84 end with $42=84/2$. So no proper subgroup of any group of order 84 has more than 42 elements.

3. The question asks for the largest possible size, so check that 42 is actually attained by some group of order 84.

4. Take $G={\mathbb{Z}}_{84}$. A cyclic group of order $n$ has exactly one subgroup of each order $d\mid n$. So ${\mathbb{Z}}_{84}$ has a subgroup of order 42, namely $⟨2⟩=\{0,2,4,\dots ,82\}$, which has $84/2=42$ elements and is proper.

5. The bound 42 is attained, so the largest possible size of a proper subgroup of a group on 84 elements is $42$.