+7 votes
3.3k views
Let $G$ be a finite group on $84$ elements. The size of a largest possible proper subgroup of $G$ is _____

edited | 3.3k views
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42.....
+2

## 4 Answers

+19 votes
Best answer
Order of a Subgroup always divides the order of Group.
Proper Subgroup of Group having order $84$ would have order $1, 2, 4, 21, 42$.

So largest order would be $42$.
by Veteran (60k points)
edited
0
group also itself called subgroup then why 84 not true?
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group also itself called subgroup but that considered as trivial subgroup. Proper subgroup is a subgroup which is not trivial . Also a group consist of only identity element is considered as trivial subgroup.
+9

Divisors of the Positive Integer 84. 1, 2, 3, 4, 67, 12, 14, 21, 28, 42, 84

All are possible proper subgroups excluding the size of 1 and 84 because they are trivial subgroups but here the question is asking, Largest possible proper subgroup possible is 42...

0

The size of the smallest proper subgroup of $G=2?$

+1
Yes it is the smallest non trivial group
+12 votes

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

by Active (1.3k points)
0
got same
0
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
+4 votes
Order of Group must be divisible by order of subgroup = 42.
by Veteran (62k points)
+1 vote
42 is correct
by (333 points)
Answer:

+36 votes
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