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A group has 11 elements. The number of proper sub-groups it can have is

1. 0
2. 11
3. 5
4. 4

Lagrange's Theorem :

The order of a finite group is a multiple of the order of its every subgroup

Here Order of group is 11 which is Prime no. So only 2 subgroups are possible.
a. Group itself (i.e. Order 11)
b. Group with order 1

but both of above are Trivial Sub Group i.e. no Proper Sub Group possible from Prime Order Group.

Group with order 1

means only identity element is there. right?

Let G be a group of prime order. Then G has no subgroups and hence is cyclic.

https://crypto.stanford.edu/pbc/notes/group/lagrange.html

but i can have 11 subgroups each containing only 1 element (the elements will be the 11 elements of the group)

like if group O=(1,2,3) then subgroup is H=(1), H=(2) , H=(3)

but i m in doubt that whether these subgroups will satisfy group property or not

@Digvijay Pandey

Order/cardinality of the group = 11.

Lagrange's theorem states that the order of every subgroup H of a group G, divides the order of G.

So what divides 11? 1 and 11.

Of order 1, there's only one subgroup, ie, {e} (identity element). This is the trivial subgroup. Every group has it.

Of order 11, the same group itself.

So, essentially there are no subgroups.

Option A

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