Groups.
[20/0%]Prove that the intersection of any family of subgroups of a group is a subgroup.
A group is said to be generated by a subset of its elements, and one writes , if the intersection of all subgroups containing coincides with (in other words, has no proper subgroup containing ). Show that in the case every element has the form , where either , or .
Show that commuting elements of an arbitrary group , having coprime orders , generate in a cyclic subgroup of order .
Show that if is a monoid generated by a set , and every element is invertible in , then is a group.
A group is a monoid in which equations of the form are uniquely solvable for any . Prove this statement.
Show that the set of the so-called affine transformations of the real line forms a group with the multiplication law . The group contains the subgroup , which fixes the point , and the subgroup of "pure translations" .
The group contains the elements of orders 4 and 3 respectively. Show that is an infinite cyclic subgroup in . Thus, the product of two elements of finite order in a group need not be an element of finite order. What is the situation in an abelian group?
Prove that a group of even order necessarily contains an element of order 2.
Prove that .
Prove that .
Prove that the alternating group , is generated by cycles of length 3, and in fact
Prove that the -th power of the cycle is a product of independent cycles, each of which has length .
Show that the order of a permutation (the order of the cyclic subgroup ) equals the least common multiple of the lengths of the independent cycles occurring in the decomposition of .
Let and for some integer . Is it true that ?
Prove that a nonempty subset of a finite (multiplicative) group is a subgroup if is closed under multiplication. Hence, in this case the requirements that contain an identity element and an inverse for each are superfluous.
What system of generators can be proposed for the multiplicative group of positive rational numbers?
Does there exist a finite system of generators in ?
Prove that, up to isomorphism, there exists only a finite number of groups of a given order .
Using Exercise 10, show that every finite group can be embedded (i.e., there exists a monomorphism for it) into a finite group with two generators.
Try to verify that the diagram (Fig. 17) depicts all subgroups of the alternating group . The symbol denotes the so-called
Fig. 17 four-group (or Klein group) , and next to the other vertices of the diagram the generators of the cyclic subgroups are indicated.
Show that all groups of order 4 are abelian and, up to isomorphism, are exhausted by the permutation groups , or else by the matrix groups:
Write out explicitly the isomorphisms .