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2.118 Prove that a ﬁnite p group G is simple if and only if G  = p .
Solution. Absent.
2.119 Show that S4 has a subgroup isomorphic to D8 .
Solution. In Example 2.139, we have seen that D8 can be viewed as certain
permutations of the 4 vertices of a square.
2.120 Prove that S4 /V ∼ S3 .
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Solution. S4 /V is a group of order 24/4=6, hence Proposition 2.135 shows
that it is isomorphic to either S3 or I6 . But S4 /V is not abelian: for example,
(1 2)V(1 3)V = (1 3)V(1 2)V because
(1 2)(1 3)[(1 3)(1 2)]−1 = (1 2)(1 3)(1 2)(1 3) = (2 3) ∈ V.
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Therefore, S4 /V ∼ S3 .
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2.121
(i) Prove that A4 ∼ D12 .
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Solution. A4 has no element of order 6, while D12 does have such
an element.
(ii) Prove that D12 ∼ S3 × I2 .
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Solution. We may suppose that D12 = a , b , where a 6 = 1 = b2
and bab = a −1 . We know that the subgroup a has order 6, hence
index 2, and so there are two cosets:
D12 = a ∪ b a .
Thus, every element x ∈ D12 has a unique factorization x = bi a j ,
where i = 0, 1 and 0 ≤ j < 6. Deﬁne H = a 2 , b ; now H ∼ S3 ,
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for it is a nonabelian group of order 6; note that H D12 because it
has index 2. If we deﬁne K = a 3 , then  K  = 2 and K D12 : it
sufﬁces to prove that aa 3 a −1 ∈ K (which is, of course, obvious)
and ba 3 b ∈ K ; but ba 3 b = a −3 = a 3 ∈ K . It is plain that
H ∩ K = {1} and H K = D12 , and so D12 ∼ H × K ∼ S3 × I2 ,
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by Proposition 2.127.
2.122 (i) If H is a subgroup of G and if x ∈ H , prove that
C H (x ) = H ∩ C G (x ).
Solution. C G (x ) = {g ∈ G : gx = xg } while C H (x ) = {g ∈ H :
gx = xg }. It follows easily that each of C H (x ) and H ∩ C G (x )
contains the other. (ii) If H is a subgroup of index 2 in a ﬁnite group G and if x ∈ H ,
prove that either x H  = x G  or x H  = 1 x G , where x H is the
2
conjugacy class of x in H . ...
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Full Document
 Fall '11
 KeithCornell
 Cyclic group, Coset, Group isomorphism, Conjugacy class, S4 /V

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