Tour:Hard nuts three (beginners)
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This page is a Hard nuts page, part of the Groupprops guided tour for beginners (Jump to beginning of tour)
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Contents
Some look-ahead hard problems
Most of the problems here are look-ahead problems: they may be hard now, but will be clear in later parts (by part six or seven), once more related material has been covered. By attempting these problems, you may find it easier to cope with the later parts.
Index and subgroups
- Suppose . Construct a natural map from the left coset space to the left coset space . Use this map to prove that . For full proof, refer: Index is multiplicative
- Prove that if are subgroups, then there is a natural bijective map between the coset space and the coset space . Here, is understood to be the set of cosets of of the form . For full proof, refer: Product formula
- Prove the product formula: If are finite subgroups, then .
- Prove that if are subgroups, then . For full proof, refer: Index satisfies intersection inequality
- Prove that if are subgroups such that and are finite and relatively prime, then .
- Prove that if a group is the union of three proper subgroups, then they all must have index two, and the intersection of any two of them must be contained in the third.
Left and right cosets
- Prove that a subset of a group occurs as the left coset of a subgroup if and only if it occurs as the right coset of a subgroup.
- Prove that if a group is the union of finitely many cosets of subgroups, then at least one of those cosets is of a subgroup of finite index.
Other hard problems
- A subgroup of a group is termed maximal among Abelian subgroups if is an Abelian subgroup, and there is no Abelian subgroup of properly containing . Prove that every group (possibly infinite) is the union of subgroups that are maximal among Abelian subgroups. For full proof, refer: Every group is a union of maximal among Abelian subgroups
- A group is termed finitely generated if it has a generating set of finite cardinality. Prove that any finitely generated group is countable. Prove that for any infinite cardinal , a group has cardinality if and only if it has a generating set of cardinality .
The murky world of cosets
These are, for the most part, insignificant problems and are included mainly for the fun element.
- (Related to the symmetric group on three letters; skip if you are not familiar with this group) Prove that the symmetric group on three letters can be expressed as the union of three disjoint cosets, each being a left coset of a different subgroup. Generalize this to dihedral groups.
- (Related to the elementary Abelian group of order eight; skip if you are not familiar with this group) Partition the elementary Abelian group of order eight into four subsets, each of which is the coset of a different subgroup. (Hint: This is equivalent to taking the vertices of the unit cube and partitioning them into pairs such that for no two pairs are the lines joining them parallel).
This page is a Hard nuts page, part of the Groupprops guided tour for beginners (Jump to beginning of tour)
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