Download ABSTRACT ALGEBRA 1, LECTURE NOTES 6: KERNELS, IMAGES

ABSTRACT ALGEBRA 1, LECTURE NOTES 6: KERNELS, IMAGES, AND
COKERNELS.
ANDREW SALCH
1. Kernels, images, and cokernels.
Definition 1.1. Let G, H be groups, written multiplicatively, and let f : G → H be a
homomorphism of groups.
• The kernel of f , written ker f , is the set of elements g ∈ G such that f (g) = 1.
• The image of f , written im f , is the set of elements of H of the form f (g) for some
g ∈ G.
Proposition 1.2. Let G be a group, written multiplicatively, and let g ∈ G. Then the
inverse element g−1 is unique, that is, there exists only one element g−1 ∈ G such that
g−1 g = gg−1 = 1.
Proof. Let u, v be two elements such that gu = ug = 1 and such that gv = vg = 1. Then:
u = u1
= ugv
= 1v
= v.
Lemma 1.3. Let G, H be groups, written multiplicatively, and let f : G → H be a homomorphism of groups. Then, for each g ∈ G, we have f (g−1 ) = f (g)−1 .
Proof.
f (g−1 ) f (g) = f (g−1 g)
= f (1)
= 1, and
f (g) f (g ) = f (gg−1 )
−1
= f (1).
−1
Hence f (g ) has the defining property of the inverse f (g)−1 .
Lemma 1.4. Let G be a group and let S be a subgroup of G. Then S is a normal subgroup
of G if and only if, for all g ∈ G, we have g−1 S g ⊆ S .
Proof. By the definition of normality, if S is normal then g−1 S g = S so clearly then
g−1 S g ⊆ S .
The converse is the part where there’s anything to check: if g−1 S g ⊆ S for all g ∈ G,
we need to show that g−1 S g = S for all g ∈ G. Suppose h ∈ G and suppose that x ∈ S
Date: September 2016.
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ANDREW SALCH
be an element which is not in h−1 S h. Since x = h−1 (hxh−1 )h, this tells us that hxh−1
cannot be in S (because if hxh−1 ∈ S , then x = h−1 (hxh−1 )h ∈ h−1 S h). But this means
that (h−1 )−1 S (h−1 ) is not contained in S , i.e., there exists an element g (namely, g = h−1 )
such that g−1 S g is not contained in S , contradicting our assumption that g−1 S g ⊆ S for
all g. Hence the assumption that g−1 S g ⊆ S for all g ∈ G implies that g−1 S g = S for all
g ∈ G.
Proposition 1.5. Let G, H be groups, written multiplicatively, and let f : G → H be a
homomorphism of groups. Then the image of f is a subgroup of H, and the kernel of f is
a normal subgroup of G.
Proof. First let’s show that im f is a subgroup of H. Since f (1) = 1 by the definition of
a group homomorphism, the unit element of H is contained in im f . Given two elements
f (g), f (g0 ) ∈ im f , we have f (gg0 ) = f (g) f (g0 ), so the product f (g) f (g0 ) in H is contained
in im f , so im f is closed under multiplication. Given an element f (g) ∈ im f , we have
f (g)−1 = f (g−1 ) by Lemma 1.3, so im f is closed under inverses. So im f is a subgroup
of H.
Now let’s show that ker f is a subgroup of G. Since f (1) = 1, ker f contains the unit
element of G. If g, g0 ∈ ker f , then f (gg0 ) = f (g) f (g0 ) = 1 · 1 = 1, so gg0 ∈ ker f , so ker f
is closed under multiplication. If g ∈ ker f , then f (g−1 ) = f (g)−1 = 1−1 = 1, so ker f is
closed under inverses. So ker f is a subgroup of G.
Now let’s show that ker f is a normal subgroup of G. Given an element x ∈ G and an
element g ∈ ker f , we have
f (x−1 gx) = f (x−1 ) f (g) f (x)
= f (x−1 )1 f (x)
= f (x)−1 f (x)
= 1,
−1
−1
so x gx ∈ ker f . So x (ker f )x ⊆ ker f . So ker f is normal, by Lemma 1.4.
Definition 1.6. Given a group homomorphism f : G → H, we define the cokernel of f ,
written coker f , as the quotient group H/ im f . (This makes sense only because we just
showed, in Proposition 1.5, that im f is a normal subgroup of G!)