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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?

1. Both $G$ and $H$ are always cyclic
2. $G$ may not be cyclic, but $H$ is always cyclic
3. $G$ is always cyclic, but $H$ may not be cyclic
4. Both $G$ and $H$ may not be cyclic

### 2 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.

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

## 4 Answers

Best answer

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 $1,2,3,6$ but $6$ is not possible as we need a proper sub-group 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

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

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### 3 Comments

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

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 )

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

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

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### 1 comment

Now its more clear sir,thanks
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.
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### 1 comment

perfect!

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.

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