GATE CSE 2018 | Question: 19

Question

Let $G$ be a finite group on $84$ elements. The size of a largest possible proper subgroup of $G$ is _____

Comments

42.....

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https://www.youtube.com/watch?v=TCcSZEL_3CQ

Answer 1

Order of a Subgroup always divides the order of Group.
Proper Subgroup of Group having order $84$ would have one of the order (proper factors of $84)$ $2,3, 4,6,7,12,14, 21,28, 42$.

So the largest order would be $42$.

Comments

So here trivial subgroups can be identity elements and the group itself. Right?

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Yes

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• Let $G$ be a finite group on $84$ elements. The size of a largest possible $\text{proper}$ subgroup of $G$ is $\color{Red}42$
• Let $G$ be a finite group on $84$ elements. The size of a largest possible subgroup of $G$ is $\color{Red}84$ (Group itself) 

Answer 2

Lagrange's theorem states that order of every subgroup of G, it must be the divisor of G.

So the largest subgroup will be 84 which is trivial, but in the question it is asking for the proper subgroup hence it will be 42.

Reference: https://en.wikipedia.org/wiki/Subgroup

Comments

got same

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but for prime divisor we  are sure  that it would be proper subgroup for other divisor it may or not be proper subgroup converse of lagrange's theorem is not true

Answer 3

Order of Group must be divisible by order of subgroup = 42.

Answer 4

42 is correct

Answer 5

Lagrange's Theorem  If H is subgroup of Finite Group G then |H|  divides |G| . So |H| will be a factor of |G|.

In the question it is asking  for proper subset so H not equal to G.

The largest factor (not equal to 84) is 42.

So Answer is 42

Answer 6

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