28.7 SUBDIVIDING A GROUP
(i) the set of elementsH′inG′that are images of the elements ofGforms a
subgroup ofG′;
(ii) the set of elementsKinGthat are mapped onto the identityI′inG′forms
a subgroup ofG.
As indicated in the previous section, the subgroupKis called thekernelof the
homomorphism.
To prove (i), supposeZandWbelong toH′, withZ=X′andW=Y′,where
XandYbelong toG.Then
ZW=X′Y′=(XY)′
and therefore belongs toH′,and
Z−^1 =(X′)−^1 =(X−^1 )′
and therefore belongs toH′. These two results, together with the fact thatI′
belongs toH′, are enough to establish result (i).
To prove (ii), supposeXandYbelong toK;then
(XY)′=X′Y′=I′I′=I′ (closure),
I′=(XX−^1 )′=X′(X−^1 )′=I′(X−^1 )′=(X−^1 )′
and thereforeX−^1 belongs toK. These two results, together with the fact thatI
belongs toK, are enough to establish (ii). An illustration of this result is provided
by the mapping Φ of →U(1) considered in the previous section. Its kernel
consists of the set of real numbers of the form 2πn,wherenis an integer; it forms
a subgroup ofR, the additive group of real numbers.
In fact the kernelKof a homomorphism is anormalsubgroup ofG.The
defining property of such a subgroup is that for every elementXinGand every
elementYin the subgroup,XY X−^1 belongs to the subgroup. This property is
easily verified for the kernelK,since
(XY X−^1 )′=X′Y′(X−^1 )′=X′I′(X−^1 )′=X′(X−^1 )′=I′.
Anticipating the discussion of subsection 28.7.2, the cosets of a normal subgroup
themselves form a group (see exercise 28.16).
28.7 Subdividing a group
We have already noted, when looking at the (arbitrary) order of headings in a
group table, that some choices appear to make the table more orderly than do
others. In the following subsections we will identify ways in which the elements
of a group can be divided up into sets with the property that the members of any
one set are more like the other members of the set, in some particular regard,