Math 4124
Monday, February 28
February 28, Ungraded Homework
Exercise 3.5.11 on page 111
Prove that
S
4
has no subgroup isomorphic to
Q
8
.
In
Q
8
we have the relation
i j
=
k
, and of course
i
,
j
,
k
all have order 4; in other words we have
the product of two elements of order 4 giving an element of order 4. Therefore if
Q
8
was
isomorphic to a subgroup of
S
4
, we would have in
S
4
the product of two elements of order
4 giving an element of order 4. Since an element of order 4 in
S
4
must be a 4cycle, which
is an odd permutation, we would now have the product of two odd permutations giving an
odd permutation. But the product of two odd permutations is always an even permutation,
so we have a contradiction. Therefore
Q
8
cannot be isomorphic to a subgroup of
S
4
.
Let
G
=
D
8
, the dihedral group of order 8. Prove that
G
has exactly one normal subgroup of
order 2, and that there are exactly 5 normal subgroups containing this subgroup.
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 Spring '08
 Staff
 Math, Algebra, Group Theory, Abelian group, gGa g−1

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