###### Question 14345 – DSSSB PGT 2018 Female

April 29, 2024###### Question 14240 – Engineering-Mathematics

April 29, 2024# Question 14236 – Engineering-Mathematics

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?

Correct Answer: D

Question 6 Explanation:

If ‘G’ is a group with sides 6, its subgroups can have orders 1, 2, 3, 6.

(The subgroup order must divide the order of the group)

Given ‘H’ can be 1 to 6, but 4, 5 cannot divide ‘6’.

Then ‘H’ is not a subgroup.

G can be cyclic only if it is abelian. Thus G may or may not be cyclic.

The H can be cyclic only for the divisors of 6 and H cannot be cyclic for any non divisors of 6.

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

Both G and H are always cyclic.

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

Both G and H may not be cyclic.

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