There's a theorem to the effect that every group of prime order is cyclic. Since 5 is prime, the assertion in the question follows from the said theorem.
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By LaGrange's Thm., the order of an element of a group must divide the order of the group. Since 3 is prime, up to isomorphism, the only group of order three is {1,x,x^2} where x^3=1. Note that this is a finite cyclic group. Since all cyclic groups are abelian, because they can be modeled by addition mod an integer, the group of order 3 is abelian.
Yes. The only group of order 1 is the trivial group containing only the identity element. All groups of orders 2 or 3 are cyclic since 2 and 3 are both prime numbers. Therefore, any group of order less than or equal to four must be a cyclic group.
Prove that if it were true then there must be a contradiction.
A 15 yr. old must be able to prove that they can find a job, live on there own, find a living area, and prove that the will no longer need there parents. And you must go to court
You must be able to take care of yourself and your child on your own with no help from government contributions or other people. You must be able to prove this in front of a Judge.