GATE CSE 2021 | Set 1 | Question: 34

Question

Let $G$ be a group of order $6$, and $H$ be a subgroup of $G$ such that $1<|H|<6$. Which one of the following options is correct?

• A. Both $G$ and $H$ are always cyclic
• B. $G$ may not be cyclic, but $H$ is always cyclic
• C. $G$ is always cyclic, but $H$ may not be cyclic
• D. Both $G$ and $H$ may not be cyclic

Comments

I think it should be option B, group of order less than 5 are always cyclic, group of order 6 may or may not be cyclic.

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i think this property is for abelian group , group of order 4 may be cyclic

http://math.mit.edu/~mckernan/Teaching/12-13/Spring/18.703/l_4.pdf

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Prime order group is Cyclic.

Order-$6$ group may or may not be cyclic. So, $G\;may\;not\;be\;cyclic.$

Given $1<|H|<6$,  According to $Lagrange’s\;Theorem:$ Order of every subgroups of $G$ divides order of $G$ 

So, possible orders of subgroup $H=\{1,2,3,6\}$. But for the given condition it should be $\{2,3\}$

$2,3$ are prime numbers. So, $H\;is\;always\;cyclic$

2. $G$ may not be cyclic, but $H$ is always cyclic.

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by langrange's theorem,

order of subgroup divides order of group.

so order of subgroup can be 2 or 3

now there is a property which states that every group of prime order is a cyclic group.

so h is cyclic but nothing cant be said about g.

Answer 1

Given $G$ is a group of order $6.$

Lagrange's Theorem: The order of every subgroup of $G$ divides the order of $G$

So, subgroup $(H)$ of $G$ can be the order of $1,2,3, \text{and }6.$ But $6$ is not possible as we need a proper sub-group, and $1$ is not possible, as mentioned in the question.

Now, any group of prime order is cyclic. If the order is not-prime both cyclic and acyclic groups are possible.

Order of $G  = 6 –$ Not a prime number so G may or may not be cyclic.

Order of $H = \{1,2,3\}  – 2$ and $3$ are prime numbers and group with 1 element is cyclic. So, H is always cyclic.

Ans: (B) $G$ may not be cyclic, but $H$ is always  cyclic

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More Read: Exactly $2$ non-isomorphic groups of order $6$ are there – cyclic or not?

Comments

But consider subgroup of size 4 and 4 is not prime, what in that case?

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A subgroup of order 4 not possible.

Given G is a group of order 6.

Lagrange's Theorem: The order of every subgroup of G divides the order of G.((https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory))

So Subgroup of G can be the order of 1,2,3.

(Full detailed ans updated )

 

 

 

 

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subgroup cant be of order 4 because subgroup should divide group, but 4 doesnt divide 6

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@Bhargav D Dave 6 but the order of H can’t be 1 bcz it is mentioned in the question.

Answer 2

Ans is B     G may not be cyclic H is always cyclic

since order of G is 6  its subgroup may have order 1,2,3,6 (according to Lagrange theorem order of subgroup must divide order of group)

now H is one of its subgroup  with condition 1<|H|<6 so H may be of order 2 or 3  which is prime 

Property of  group says if a group has prime order it is cyclic so H must be cyclic

Order of G is 6  (not prime ) so it may or may not  be cyclic 

Comments

Now its more clear sir,thanks

Answer 3

For answering this question we need to know mainly two theorems:

$\text{(1) Lagrange’s Theorem : If $H$ is a subgroup of some finite group $G$ then }$

$\text{$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$the order of $G$ is divisible by the order of $H$.}$

$\text{(2) If a group has a prime order then it is cyclic.}$

Now it is given that $G$ is a group of order $6$. Since $6$ is not a prime number, so $G$ may or may not be cyclic [Since Theorem $(2)$ says being prime is just a sufficient condition for a group to be cyclic but not necessary]

Also, it is that that order of $H$ can be $2,3,4,5$ [as per the question $1<|H|<6$]. Now Theorem $1$ says that it is necessary for the order of a group to be divisible by the order of the subgroup.

Hence the only feasible orders of $H$ are $2,3$. [The orders $4,5$ are ruled out since they do not divide $6$.]

Now what we see is that the possible orders of $H$ are prime and hence from Theorem $2$, $H$ has to be cyclic.

So option $(B)$ is the correct option.

Comments

perfect!

Answer 4

option A is wrong.

Option B, if H is cyclic and is proper subgroup then G may or may not be cyclic.  If H is cyclic and is improper subgroup then G is always cyclic. Hence considering H to be proper subgroup, option B is correct.

Option C is wrong.

Option D is wrong.

Answer 5

By Lagrange's Theorem , in question it is already given that H is subgroup therefore order of H will divide order of G , 

|G| = 6 therefore |H| = 2,3 we know order 2 , 3 have only one template and bothe of them cyclick and abelian . credit @deepak poonia  sir @GO Classes

Answer 6

Analysis

 

1. Analysis of the Subgroup H

 

• According to Lagrange's Theorem, the order (number of elements) of a subgroup H must divide the order of the group G.

• The order of G is 6. The divisors of 6 are {1, 2, 3, 6}.

• The problem adds the constraint that 1 < |H| < 6.

• Combining these, the only possible orders for the subgroup H are 2 or 3.

• A key theorem in group theory states that any group of a prime order is always cyclic.

• Since both 2 and 3 are prime numbers, the subgroup H is always cyclic, regardless of its specific structure.

 

2. Analysis of the Group G

 

• The order of G is 6, which is a composite number (not prime).

• A group of composite order may or may not be cyclic.

• To prove that G is not always cyclic, we just need to find one example of a non-cyclic group of order 6.

• The Symmetric Group S₃ (the group of all permutations of 3 elements) is a classic example. It has 3! = 6 elements and is non-cyclic.

• Since there exists at least one non-cyclic group of order 6, we cannot guarantee that G will always be cyclic. Therefore, G may not be cyclic.

 

Conclusion

 

• H is always cyclic.

• G may not be cyclic.

This matches the statement in option B.