GATE CS 2021 Set 1 Q.44 (CS Q.34): Monoids and Groups

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?

Both $G$ and $H$ are always cyclic.

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

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

Both $G$ and $H$ may not be cyclic.

Official answer and solution

Official answer: B
From: Official GATE 2021 answer key

Solution

Apply Lagrange's theorem on subgroup orders and recall properties of groups of prime order.

Let $G$ be a group of order $|G| = 6$ . By Lagrange's theorem, the order of any subgroup $H$ of $G$ must divide $|G| = 6$ . The divisors of 6 are 1, 2, 3, and 6.

We are given $1 < |H| < 6$ , so $|H|$ must be either 2 or 3. Both 2 and 3 are prime numbers.

A fundamental theorem of group theory states that every group of prime order is cyclic. Therefore, every proper subgroup $H$ of order 2 or 3 is always cyclic.

On the other hand, up to isomorphism there are two distinct groups of order 6: the cyclic group $\mathbb{Z}_6$ , and the symmetric group $S_3$ (dihedral group $D_3$ ). $S_3$ is non-abelian and therefore not cyclic. Thus, $G$ may not be cyclic.

Hence, $G$ may not be cyclic, but $H$ is always cyclic (option B).

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question: 44 of 65, printed as CS Q.34
Section: Computer Science and Information Technology

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