10 = 2*5, a product of 2 primes, and 2 divides (5-1). So there are only two groups.
There are 5 groups of order 8 up to isomorphism. 3 abelian ones (C8, C4xC2, C2xC2xC2) and 2 non-abelian ones (dihedral group D8 and quaternion group Q)
Since 121 is the square of a prime, there are only two distinct isomorphic groups.
If the set is of finite order, that is, it has a finite number of elements, n, then the number of subsets is 2n.
Yes, you divide the number of expected outcomes by the number of possible outcomes in order to determine probability.
No. As long as there are more than two, the number's composite.
There are 5 groups of order 8 up to isomorphism. 3 abelian ones (C8, C4xC2, C2xC2xC2) and 2 non-abelian ones (dihedral group D8 and quaternion group Q)
There are two: the cyclic group (C10) and the dihedral group (D10).
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.
7 groups, use the structure theorem
Since 121 is the square of a prime, there are only two distinct isomorphic groups.
Number of generators of that group
A group is when you have a number of people or things together. In order for it to be a group it has to be 3 or more.
A group is when you have a number of people or things together. In order for it to be a group it has to be 3 or more.
The order of a cyclic group is the number of distinct elements in the group. It is also the smallest power, k, such that xk = i for all elements x in the group (i is the identity).
The number of protons determine what element it is, the number of neutrons determine what isotope it is.
No
If the set is of finite order, that is, it has a finite number of elements, n, then the number of subsets is 2n.