Math 4124
Wednesday, March 30
Seventh Homework Solutions
1. Exercise 4.4.3. Prove that under any automorphism of
D
8
,
r
has at most two possible
images and
s
has at most four possible images. Deduce that

Aut
(
D
8
)
 ≤
8. Here
D
8
=
{
r
,
s

r
4
=
s
2
=
1
,
rs
=
sr

1
}
.
As usual,
D
8
=
{
1
,
r
,
r
2
,
r
3
,
s
,
sr
,
sr
2
,
sr
3
}
. Here

r

=

r
3

=
4 and

r
2

=

s

=

sr

=

sr
2

=

sr
3

=
2. Let
α
∈
Aut
D
8
. Since

r

=
4, we see that

α
r

=
4 (the order
of an element is preserved by an isomorphism, so in particular it is preserved by an
automorphism), hence
α
r
=
r
or
r
3
; in other words there are at most two possible
choices for
α
r
.
Next

s

=
2, hence

α
s

=
2, so it would appear at ﬁrst sight that there are 5 possible
choices for
α
s
, namely
r
2
,
s
,
sr
,
sr
2
,
sr
3
. However, we can rule out the case
α
s
=
r
2
.
Indeed we have
α
r
=
r
or
r
3
(from the ﬁrst paragraph), hence
α
r
2
= (
α
r
)
2
=
r
2
or
r
6
and we see that
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 Spring '08
 Staff
 Math, Algebra, Normal subgroup, Subgroup, Cyclic group, Sylow 19subgroups, Sylow 5subgroups

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