Hopf's formula for Schur multiplier
Let be a group isomorphic to the quotient group , where is a free group and is a normal subgroup of . Then, the Schur multiplier of , denoted , which is the same as the second homology group for the trivial group action of on the integers, denoted , is an abelian group given by the formula:
In other words, it equals the quotient of the intersection of with the commutator subgroup of by the focal subgroup of in . ( equals the focal subgroup because is a normal subgroup of ).
Note that any choice of generating set for gives a choice of and for which the theorem can be applied: is the free group on those generators with the natural surjection, and is the kernel of the surjection.
- The definition of Baer invariant generalizes this formula
- Hopf's formula for nilpotent multiplier
- Variant of Hopf's formula for Schur multiplier for nilpotent group that uses the free nilpotent group of class one more
- Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup
- Schur multiplier of free group is trivial
Consider the short exact sequence:
This gives a related short exact sequence:
Proof of being an initial object in the category of central extensions with homoclinisms
On account of being free, this short exact sequence gives an initial object in the category of central extensions with quotient group . Explicitly, for any central extension:
there is a (unique) homomorphism that, composed with the quotient map , gives the quotient map . Here is a sketch of the process:
- Consider a freely generating set for . Take its image in .
- For each element in the generating set for , pick an element of that maps to its image in .
- Consider the homomorphism obtained from the set map given from the generating set of above. Note that this exists because is free.
- Verify that is in the kernel of , so descends to a map , which restricts to a homomorphism .
Note that the second step of the construction introduces non-uniqueness. However, from general considerations, the map has to be unique, so this is a non-issue.
Consequence for the exterior square and Schur multiplier
This immediately implies that and (the Schur multiplier of ) is the kernel of the quotient map . Let's calculate both:
- simplifies to . Thus, .
- The kernel of the map from to is , hence the kernel of is . The kernel of the map is the intersection . This, then, is the Schur multiplier.
Proof using the Stallings exact sequence
Given: A group written as a quotient group of a free group by a normal subgroup of .
To prove: Let and . The Schur multiplier , defined as the kernel of the commutator map homomorphism , is isomorphic to .
Proof: Consider the short exact sequence:
The corresponding Stallings exact sequence is:
Since is free, is trivial. Thus, the sequence simplifies to:
The homomorphism has kernel equal to . Thus, by exactness, that is also the image of . Since is trivial, is an isomorphism to its image. Thus, we get: