65
(iii)
Prove that ker
γ
=
Z
(
G
)
.
Solution.
Absent.
(iv)
Prove that im
γ
C
Aut
(
G
)
.
Solution.
Absent.
2.93
If
G
is a group, prove that Aut
(
G
)
={
1
}
if and only if

G
≤
2.
Solution.
Instructors are cautioned: assume that
G
is
f
nite. The exercise
is, in fact, true as stated, but it wants Zorn
’
s lemma at one stage (see the
following argument).
If there is
a
∈
G
with
a
/
∈
Z
(
G
)
, then conjugation by
a
is a nontrivial
automorphism; therefore,
G
is abelian. The map
x
7→−
x
is an automor
phism of
G
; if it is trivial, then
x
=−
x
for all
x
∈
G
. Thus, we may
assume that that
G
is an abelian group (which we now write additively)
with 2
x
=
0; that is,
G
is a vector space over
F
2
(the
f
eld with 2 ele
ments). If
G
is
f
nitedimensional and dim
(
G
)
≥
2, then any nonsingular
matrix other than the identity corresponds to a nontrivial automorphism.
For example, if
v
1
,v
2
,...,v
n
is a basis, then there is an automorphism
which interchanges
v
1
and
v
2
and which
f
xes
v
3
,...,v
n
.I
f
G
is in
f
nite,
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 Fall '11
 KeithCornell
 Vector Space, Normal subgroup, Abelian group, Solution., Quotient group, Zorn

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