Download Homework 7 October 21, 2005 Math 521 Direction: This homework

Homework 7
October 21, 2005
Math 521
Direction: This homework is due on October 28, 2005. In order to receive full
credit, answer each problem completely and must show all work.
1. Consider the group G = {1, i, −1, −i} under multiplication. For an arbitrary but fixed
element a in G, define a map ha : G → G by ha (x) = xa for all x ∈ G. Find the maps hi ,
h1 , h−1 and h−i . Show that they are permutations on the set of elements in G.
Answer: Let G be {1, i, −1, −i}. The Cayley table of G is the following:
·
1
i
−1
−i
1 i
Hence h1 =
1 i
1 i −1
hi =
−i 1 i
tations on G.
1
1
i
−1
−i
i
i
−1
−i
1
−1
−1
−i
1
i
−i
−i
1
i
−1
−1 −i
1 i −1 −i
1
i −1 −i
, hi =
, h−1 =
,
−1 −i
i −1 −i 1
−1 −i 1
i
−i
. Since these functions are one-to-one and onto, they are permu−1
2. For each a in the arbitrary group G, define a mapping ha : G → G by ha (x) = xa for all
x in G. Prove that ha is a permutation on the set of elements in G.
Answer: For each a ∈ G, define the map ha : G → G by ha (x) = xa for all x ∈ G. To show
ha is a permutation we have to show ha is well-defined, one-to-one and onto. First we show
ha is well-defined. Suppose x = y. Then xa = ya. Hence ha (x) = ha (y). Therefore ha is
well-defined. Next, we show ha is one-to-one. Suppose x 6= y. Then xa 6= ya. This implies
that ha (x) 6= ha (y). So ha is one-to-one. Finally, we show that ha is onto. Let b ∈ G and
find an element x ∈ G such that ha (x) = b. This implies that xa = b which yields x = ba−1 .
This shows that ha is onto.
3. For each a in the arbitrary group G, define a mapping ha : G → G by ha (x) = xa
for all x. Show that the set of permutations H = {ha | a ∈ G} is a group under function
composition.
Answer: Let H = { ha | a ∈ G }. We want to show that H is a group under function
composition. If e is the identity of G, then he is the identity of H. This can be seen from
the followings: he ha (x) = he (ha (x)) = he (xa) = (xa)e = xa = ha (x). So he ha = ha for any
a ∈ G. Similarly, we have ha he = ha for every a ∈ G. Next, we show that if a ∈ G, then ha−1
is the inverse ha . To see this consider ha−1 ha (x) = ha−1 (xa) = (xa)a−1 = x = xe = he (x).
Hence ha−1 ha = he and similarly ha ha−1 = he . Therefore ha−1 is the inverse ha in H. Since
the function composition is associative, the associativity law holds in H. Thus H is a group
consisting of a set of all permutations on G.
4. For each a in the arbitrary group G, define a mapping ha : G → G by ha (x) = xa for all
x in G. Show that the permutation group H = {ha | a ∈ G} is anti-isomorphic to the group
G. [Note: A function φ : H → G is an anti-isomorphism if and only if φ is one-to-one,
onto, and satisfies φ(xy) = φ(y)φ(x) for all x, y ∈ H.]
Answer: We want to show G is anti-isomorphic to H. That is, we want to produce an
anti-isomorphism from G to H. Consider the mapping φ : G → H defined by φ(a) = ha for
all a ∈ G. First, we show φ is well-defined. Suppose a = b. This implies xa = xb for x ∈ G.
Hence from the definition of ha , we get ha (x) = hb (x) for all x ∈ G. This yields ha = hb and
therefore φ(a) = φ(b). Hence we see that φ is well-defined. Second, we show that φ is oneto-one. Suppose a 6= b. Then xa 6= xb. Hence ha (x) 6= hb (x) and thus ha 6= hb . Therefore
φ(a) 6= φ(b). That is φ is one-to-one. Third, we show that φ is onto. Since φ(a) = ha for
each a ∈ G, φ is clearly onto. Finally, we show that φ(ab) = φ(b)φ(a) for all a, b ∈ G. For
this first we show hab = hb ha . Since hab (x) = xab = (xa)b = hb (xa) = hb (ha (x)) = hb ha (x),
we see that hab = hb ha Since φ(ab) = hab = hb ha = φ(b)φ(a), the mapping φ is an antiisomorphism and G and H are anti-isomorphic to each other.
5. For each a in an arbitrary group G, define a mapping φa (x) = axa−1 for all x ∈ G. Prove
that φa is an isomorphism from G onto G.
Answer: Define φa : G → G by φa (x) = axa−1 for all x ∈ G. First, we show φa is welldefined. Suppose x = y. This implies ax = ay and axa−1 = aya−1 . Hence φa (x) = φa (y).
Therefore φa is well-defined. Second, we show φa is one-to-one. Suppose x 6= y. Then
axa−1 6= aya−1 and hence φa (x) 6= φa (y). So φa is one-to-one. Third, we show φa is onto.
Pick an arbitrary element b in G and find an element x in G such that φa (x) = b. If an
element x exists, then axa−1 = b. Hence x = a−1 ba. Since a−1 ba ∈ G, such an element x
always exists. Thus φa is onto. Finally, we show φa preserves group operation. Consider
φa (xy) for all x, y ∈ G. Since φa (xy) = axya−1 = axa−1 aya−1 = φa (x)φa (y). Therefor φa
is an isomorphism from G onto G. [ Recall that φa is known as the inner automorphism of
the group G.]
6. Let G be a cyclic group of order 105. Find all subgroups of G.
Answer: Since G is a cyclic group of order 105, therefore G =< g > for some g ∈ G, and
G ' Z105 . The divisors of 105 are 1, 3, 5, 7, 15, 21, 35, 105. Hence the subgroups of G are:
< g 105 >, < g 35 >, < g 21 >, < g 15 >, < g 7 >, < g 5 >, < g 3 > and < g 1 >. Note that
< g 105 > is same as < e >.
7. Find an isomorphism from the group of integers (Z, +) to the group of even integers
(2Z, +).
Answer: We want to find an isomorphism from φ : Z → 2Z. Define φ(x) = 2x for each
x ∈ Z. Then φ is one-to-one and also onto. Next show that φ(x + y) = φ(x) + φ(y) which
is really easy. Hence Z ' 2Z.
8. Show that U (8) is isomprphic to U (12).
Answer: We want to show U (8) ' U (12). Since U (8) = {1, 3, 5, 7} and U (12) = {1, 5, 7, 11}
define a mapping φ : U (8) → U (12) by
φ=
1 3
1 5
5 7
7 11
.
Then clearly π is one-to-one and onto. Next we show φ(xy) = φ(x)φ(y) for all x, y ∈ U (8).
U (8)
1
3
5
7
1
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
7
5
3
1
U (12)
1
5
7
11
1
1
5
7
11
5
5
1
11
7
7
7
11
1
7
11
11
7
5
1
We see that if we replace each entry x in the multiplication table for U (8) by φ(x), then we
get the multiplication table for U (12). Hence φ satisfies φ(xy) = φ(x)φ(y) for all x, y ∈ U (8).
9. Let C be the set of complex numbers and
M=
a −b
b a
| a, b ∈ R .
Show that C and M are isomorphic under addition.
Answer: We want to show that C ' M . Define a function φ : C → M as
φ(a + ib) =
a −b
b a
for all a + ib ∈ C.
First, we show that φ is well-defined. Suppose a1 + ib1 = a2 + ib2 . Then we have a1 = a2
a1 −b1
a2 −b2
and b1 = b2 . Therefore
=
and φ(a1 + ib1 ) = φ(a2 + ib2 ). Thus
b1 a1
b2 a2
φ is well-defined. Second, we show φ is one-to-one. Suppose φ(a1 + ib1 ) = φ(a2 + ib2 ).
a1 −b1
a2 −b2
Then
=
. Therefore a1 = a2 and b1 = b2 . This implies that
b1 a1
b2 a2
a1 + ib1 = a2 + ib2 . Hence φ is one-to-one. It is easy to see φ is onto. Finally, we show φ
satisfies φ(z1 + z2 ) = φ(z1 ) + φ(z2 ) for all z1 , z2 ∈ C. For z1 = a1 + ib1 and z2 = a2 + ib2
consider
φ(z1 + z2 ) = φ(a1 + ib1 + a2 + ib2 )
= φ(a1 + a2 + i(b1 + b2 ))
a1 + a2 −b1 − b2
=
b1 + b2 a1 + a2
a1 −b1
a2 −b2
=
+
b1 a1
b2 a2
= φ(a1 + ib1 ) + φ(a2 + ib2 )
= φ(z1 ) + φ(z2 ).
Therefore φ is an isomorphism from C onto M , and C ' M .