Madeeasy-CBT

Question

Comments

why P is not subgroup of [G,*]

for [P,*]

1) forall x,y belongs to P x*y belongs to P

2) identity element of G is in P i,e 1

this is satisfying right ??

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for Q

x = 2019k +1 = 2019*0 + 1= 1

y = 2019k +1 = 2019*1 +1 =2020

x+y = 2021 ...which we cannot write in the form of 2019k + 1. where k = ...-2,-1,0,1,2,....

=>Q is not closed under subgroup H.

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For closure property it is not that if x,y belongs to SubG then x*y should belong to G

It is like , if x,y belongs to SG then x*y should belong to Subgroup

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let alone P, even G is not a group. elements are not having their inverse.

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even (G,*) is not a group. inverse does not exist for every element.

this question is itself wrong.

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The elements of $P$ are not having inverse, it should not be a subgroup.

For $Q$, it is not having the identity element.

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Here G is not a group. Inverse doesn't exists for elements of G, but question says that G is a group.

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@aambazinga exactly.