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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.
Show that an element of a group has order n if and only if it generates a cyclic group of order n?
An element ( g ) of a group ( G ) has order ( n ) if the smallest positive integer ( k ) such that ( g^k = e ) (the identity element) is ( n ). This means the powers of ( g ) generate the set ( { e, g, g^2, \ldots, g^{n-1} } ), which contains ( n ) distinct elements. Therefore, the cyclic group generated by ( g ), denoted ( \langle g \rangle ), has exactly ( n ) elements, thus it is a cyclic group of order ( n ). Conversely, if ( \langle g \rangle ) is a cyclic group of order ( n ), then ( g ) must also have order ( n ) since ( g^n = e ) is the first occurrence of the identity.
What criteria a knight must meet in order to graduate?
He must prove loyalty and mainly prove himself in battle
If PS are distinct primes show that an abelian group of order PS must be cyclic?
If ( G ) is an abelian group of order ( PS ), where ( P ) and ( S ) are distinct primes, then by the Fundamental Theorem of Finite Abelian Groups, ( G ) can be expressed as a direct product of cyclic groups of prime power order. The possible structures for ( G ) are ( \mathbb{Z}/PS\mathbb{Z} ) or ( \mathbb{Z}/P^k\mathbb{Z} \times \mathbb{Z}/S^m\mathbb{Z} ) with ( k ) and ( m ) both ( 1 ). However, since ( P ) and ( S ) are distinct primes, the only way for ( G ) to maintain order ( PS ) while being abelian is for it to be isomorphic to ( \mathbb{Z}/PS\mathbb{Z} ), which is cyclic. Thus, ( G ) must be cyclic.
In order to prove or disprove your hypothesis you must?
be testable
What must the prosecution prove in order to secure a guilty verdict?
The prosecution must prove beyond a reasonable doubt that the defendant committed the crime they are accused of in order to secure a guilty verdict.
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.
Is acetic acid aromatic?
No. There are two molecular requirements that must be met to indicate aromaticity. In order to be aromatic a molecule must: 1) have a ring of pi electrons above and below the molecule. This means: a. The molecule must be cyclic. b. All atoms in the ring must be sp2 hybridized c. The molecule must be planar. 2) have an odd number of pairs of pi electrons. Since acetic acid is not cyclic, it cannot meet the first requirement and therefore cannot be aromatic. There are, however, delocalized electrons that increase the stability of the compound.
How do you prove that the group has no subgroup of order 6?
To prove that a group ( G ) has no subgroup of order 6, we can use the Sylow theorems. First, we note that if ( |G| ) is not divisible by 6, then ( G ) cannot have a subgroup of that order. If ( |G| ) is divisible by 6, we analyze the number of Sylow subgroups: the number of Sylow 2-subgroups ( n_2 ) must divide ( |G|/2 ) and be congruent to 1 modulo 2, while the number of Sylow 3-subgroups ( n_3 ) must divide ( |G|/3 ) and be congruent to 1 modulo 3. If both conditions cannot be satisfied simultaneously, it implies that no subgroup of order 6 exists.
