Hopf's formula for Schur multiplier
Contents
Statement
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.
Related facts
- 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
Facts used
- Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup
- Schur multiplier of free group is trivial
Proof
Direct proof
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: