For example, let's see what we get from the exact sequence
where
is a positive integer, and
has the structure of
trivial
module. By definition we have
and
.
The long exact sequence begins
From the first few terms of the sequence and the fact
that
surjects onto
, we see that
on
is injective.
This is consistent with our observation above that
. Using this vanishing and the right side of the
exact sequence we obtain an isomorphism
As we observed above, when a group acts trivially the
is
, so
![$\displaystyle \H^2(G,\mathbf{Z})[m] \cong \Hom (G,\mathbf{Z}/m\mathbf{Z}).$](img1936.png) |
(11.1) |
One can prove that for any
and any module
that the group
has exponent dividing
(see Remark 11.3.4).
Thus (11.2.1) allows
us to understand
, and this comprehension arose
naturally from the properties that determine
.
William Stein
2012-09-24