Why Read This Book? - Index of

Ker �
e
x 2
x maps to
x Ker �
G
�
onto
H
x 4
x 1
x 3
Ker � x1Ker �
e x1 x 2Ker �
x 2
x 4
x 4Ker �
x 3Ker �
Figure 8.3 Isomorphism between G/ Ker(φ) and H.
f(x 1 ) 5 y(x 1Ker �)
x 3
y
8.6 Group Morphisms 285
with the epimorphism φ : G → H. Now create a carbon copy of G, only collect
all the elements of Ker(φ) and hogtie them together into a single entity in the
sketch of G/ Ker(φ). Similarly, go to each coset of Ker(φ), take all its elements,
and lump them together into a single entity in G/ Ker(φ) to create a visualization
of G/ Ker(φ). We have a link between G and G/ Ker(φ), and that is the function
that maps x ∈ G to x Ker(φ). It might be that φ is one-to-one, or maybe it is not.
But the extent to which φ collapses elements of G down to single elements of H is
precisely the extent to which elements of G clump together into cosets in G/ Ker(φ)
(Exercise 8.6.18), which itself depends on the size of the kernel. The task is to show
that G/ Ker(φ) and H are isomorphic by finding the required mapping between
them. We must map each coset in G/ Ker(φ) to an element of H, and the way to
do this is to choose a coset x Ker(φ), grab some element of it, say x, and send the
whole coset to φ(x) in H. In the following proof, we use · and ∗K to represent the
binary operations in H and G/ Ker(φ), respectively.
Proof. We must find a bijection ψ : G/ Ker(φ) → H such that
ψ[x Ker(φ) ∗K y Ker(φ)] =ψ[x Ker(φ)]·ψ[y Ker(φ)] (8.78)
for all x Ker(φ), y Ker(φ) ∈ G/ Ker(φ). So define ψ : G/ Ker(φ) → H by
ψ[x Ker(φ)] =φ(x) (8.79)