Math 455
Homework 4:
Homework 4
November 5, 2007
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Section 10. 6, 7, 16, 27, 28, 40 Section 11. 1, 10, 16, 46, 47
10.6. Find all left cosets of the subgroup { 0, 2 } of the group D4 given by Table 8.12.
Solution: Let H = { 0, 2 } = 0H = 2H
1H = 2H = { 1, 2 }
2H = 1H = { 2, 1 }
3H = 1H = { 3, 1 }
10.7. Repeat the preceeding exercise, but find the right cosets this time. Are they the same
as the left cosets?
Solution: Let H = { 0, 2 } = H0 = H2
H1 = H1 = { 1, 1 }
H2 = H1 = { 2, 1 }
H3 = H2 = { 3, 2 } Note the left coset is not equal to the right coset.
10.16. Let  = ( 1, 2, 4, 5 )( 3, 6 ) in S6. Find the index of  in S6.
Solution: Since ( 1, 2, 4, 5 ) and ( 3, 6 ) are disjoint, the order of  is the least
common multiple of the order of the disjoint cycles, i.e. the least common multiple
of 4 and 2 which is 4. So the index of  in S6 is:
( S6 :  ) = S6     = 6!  4 = 180
10.27. Let H be a subgroup of a group G and let g  G. Define a one-to-one map of H onto
Hg. Prove that your map is one-to-one and is onto Hg.
Proof: Let Hg = { hg h  H }. Let : H  Hg be defined as (h) = hg for all h  H.
We show  is one-to-one. Suppose (h1) = (h2), then h1g = h2g. Since g  G and
G is a group, g-1  gg-1 = e, so h1gg-1 = h2gg-1  h1 = h2. So  is one-to-one.
We show  is onto. For each y  Hg  h  H  y = hg by the way we defined Hg, so
 is onto.
Math 455
Homework 4
November 5, 2007
10.28. Let H be a subgroup of a group G such that g-1hg  H for all g  G and all h  H.
Show that every left coset gH is the same as the right coset Hg.
Proof: Using set notation, H = { h  g-1hg  H, g  G }
We show Hg  gH.
For some g  G, and some h  H, let hg  Hg.
But hg = ehg = gg-1hg since e = gg-1. Since g-1hg  H, gg-1hg  gH.
Since hg = gg-1hg, hg  Hg  hg  gH. So, Hg  gH
We show gH  Hg.
For some g  G, and some h  H, let gh  gH.
But gh = ghe = ghg-1g since e = g-1g. Since g = (g-1)-1, ghg-1g = (g-1)-1h(g-1)g.
Since G is a group, g  G  g-1  G  (g-1)-1h(g-1)  H  (g-1)-1h(g-1)g  Hg.
Since gh = (g-1)-1h(g-1)g, gh  gH  gh  Hg. So, gH  Hg.
Hg  gH  gH  Hg  gH  Hg.
10.40. Show that if a group with identity e has finite order n, then an = e for all a  G.
Proof:
Case 1. Suppose there does not exist x    ax = e. Then the set { ar  r   } is
not a group since it does not contain an identity element. But this contradicts Thm 5.17
which states every element of a group generates a subgroup. So the antecedent of case 1 is
false.
Case 2. Suppose x    ax = e for some a  G. Then a generates a cyclic
subgroup of G of order x. (We assume x is the smallest positive x that ax = e.) By the
Theorem of Lagrange, x divides n which implies x  n and m    mx = n. So an =
(ax)m = em = e.
11.1. List the elements of 2  4. Find the order of each of the elements. Is this
group cyclic?
Solution: 2  4 = { (0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), (1,3) }
(0,1) = (0,3) = 4
(1,1) = (1,3) = 4
(0,2) = 2
(1,0) = 2
(1,2) = 2
Since no element generates the group,
(0,0) = 1
2  4 is not cyclic.
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Math 455
Homework 4
November 5, 2007
11.10. Find all proper nontrivial subgroups of 2  2  2.
Solution:
{ (0,0,0), (0,0,1) }, { (0,0,0), (0,0,1), (0,1,0), (0,1,1) },
{ (0,0,0), (0,1,0) }, { (0,0,0), (0,1,0), (1,0,0), (1,1,0) },
{ (0,0,0), (0,1,1) }, { (0,0,0), (1,0,0), (0,0,1), (1,0,1) },
{ (0,0,0), (1,0,0) }, { (0,0,0), (0,0,1), (1,1,0), (1,1,1) },
{ (0,0,0), (1,0,1) }, { (0,0,0), (0,1,0), (1,0,1), (1,1,1) },
{ (0,0,0), (1,1,0) }, { (0,0,0), (1,0,0), (0,1,1), (1,1,1) },
{ (0,0,0), (1,1,1) }, { (0,0,0), (0,1,1), (1,0,1), (1,1,0) },
11.16. Are the groups 2  12 and 4  6 isomorphic? Why or why not?
Solution: Let the symbol  represent the phrase “is isomorphic to”.
6  2  3 and 12  3  4 each by Theorem 11.5.
So, 2  12  2  3  4, and 4  6  4  2  3  2  3  4.
Thus, 2  12  4  6.
11.46. Prove that a direct product of abelian groups is abelian.
Proof: Let (a1, a2, …, an), (b1, b2, …, bn)  in= 1 Gi, then
(a1, a2, …, an)(b1, b2, …, bn) = (a1b1, a2b2, …, anbn) Since Gi is abelian,
(a1b1, a2b2, …, anbn) = (b1a1, b2a2, …, bnan) = (b1, b2, …, bn)(a1, a2, …, an).
Thus the direct product of abelian groups is abelian.
11.47. Let G be an abelian group. Let H be the subset of G consisting of the identity e
together with all elements of G of order 2. Show that H is a subgroup of G.
Proof: By hypothesis, e  H. We show x  H has an inverse and we show H is closed.
Show Inverse: Let x  H. Then for x  e has order 2, so x2 = xx = e  x = x-1. (e = e-1.)
Show closure. Let x, y  H. Then x2 = y2 = e, and (xy)2 = (xy)(xy) = x(yx)y. Since G is
abelian, x(yx)y = x(xy)y = (xx)(yy) = x2y2 = ee = e. So, xy has order 2. Also xy  G since G
is a group. So, xy  H. Thus, H is closed.
Thus by Theorem 5.14 H is a subgroup of G.
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