every non empty subset of order n that divides the order m of a finite group is a left coset or a right coset in a quotient group. True or False
Consider the cyclic group G=Z/6Z={0,1,2,3,4,5} of order 6 and
the subset A={0;4} of order 2.
A is not a coset.
(0,4)is clearly a subset of the group Z/6Z. It's order is 2.
My question is whether this subset is a coset in any other group.
Recollect central series and normal series.
Order of (0,4) can divide the orders of many groups. So, this (0,4) with respect to the operation in that group may allow it to be a coset
Either I did not understand the question or my example works good. If you take G=Z/6Z with + operation modulo 6 then what quotients do you have? Non-trivial quotients are Z/2Z of order 2 and Z/3Z of order 3. In the first case 0 and 4 are equal mod 2, hence {0,4}={0}, in the second one {0,4} is not a coset.
And what happens if one considers, say, the simple alternating group A_5 of order 60 with no non-trivial quotients? Any subset of order 30 is not a coset.
you are right. because, the discussion is on finite groups. in the case of infinite groups, some more things are to be considered like {0, 1} is a subset of the groups (Z,+) as well as (R, +). however, the order of the subgroup dividing the order of the group is another puzzle.
By and large, your explanation will throw light onto the discussion and to the satisfaction of the reader.
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