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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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Prove that a group of order three is abelian?
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.
Is every group whose order is less than or equal to 4 a cyclic group?
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.
How do you prove a conjecture is false?
Prove that if it were true then there must be a contradiction.
How does a 15 year old become emancipated in California?
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
How can you emancipate yourself in IL?
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.
Prove that a group of order three is abelian?
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.
How do we prove that a finite group G of order p prime is cyclic using Lagrange?
Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.
Is every group whose order is less than or equal to 4 a cyclic group?
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.
Proof or Disprove 'If every proper subgroup of G is cyclic then G must be cyclic'?
No! Take the quaternion group Q_8.
What criteria a knight must meet in order to graduate?
He must prove loyalty and mainly prove himself in battle
In order to prove or disprove your hypothesis you must?
be testable
Why do you have to keep all conditions the same in the control group?
The sole purpose for a control group is so the researcher can compare his/her results with the experimental group in order to prove his/her hypothesis is correct by showing conditions before they have changed any variable. in order to achieve this and make it fair all conditions must be kept the same in order to make a fair comparison.
What must be present for fatigue crackling to occur?
cyclic stress
What is the order of an element in a group?
The order of an element in a multiplicative group is the power to which it must be raised to get the identity element.
What characteristics must a problem possess in order for it to be studied scientifically?
It must have a control group, experimental group, and a experimental variable
Do a dealer gave you the title when you buy the car cash?
In order to prove ownership you must have the title.
If a parallelogram is inscribed in a circle then it must be a?
If a parallelogram is inscribed in a circle then it must be a cyclic quadrilateral.
