SOLUTIONS

Mathematics 581   Problem Set 3   Due Monday, April 22

1. (a) Prove that if in a group G,  (ab)2 = a2b for all  a  and  b  in  G, then  G  is Abelian. (Hint: Start by writing the elements in the equation in full without using exponents.)
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For all  a, b  in  G, (ab)(ab) = aabb.  that is,  abab = aabb.  Cancel  a on the left and  b  on the right to get  ba = ab.  Therefore,  g  is Abelian.
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(b) Prove that if in a group  G, a2 = e  for all  a  in  G, then  G  is Abelian. (Hint: Show that the condition in part (a) is fulfilled.)
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a2 = e for all  a  in  G.  Hence,  b2 = e  and  (ab)2 = e.  Hence  (ab)2 =  e  = a2b2, and by part (a),  G  is Abelian.
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2. Let  G = (g)  be a finite cyclic group with n elements. Define a mapping  f  of  Zn  into  G  by
f([k]) = gk. Prove that f is well-defined.
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First,  gn = e, because  g  is a generator of  a cyclic group with  n  element.  Suppose that  [k] = [m].  in  Zn.  Then  m = nq + k.  then   gm =  gnq + k = (gn)qgk = eqgk = gk.
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3. Let  G  e a group and H  be a subgroup of  G. Prove that for any elements  a  and  b of  G, either Ha = Hb or Ha  and  Hb  are disjoint . (For some a, b,  Ha could equal Hb, while for other a, b,  the cosets Ha and Hb could be disjoint.)
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Suppose that  Ha  and  Hb  are not disjoint.  That means that they have an element  c  in common.  So,
c = h1a = h2b  for some  hand hin  H.   Multiply by  h1'  on the left to get  a = h1' h2b.  If  x  is in  Ha, then  x = ha  for an  h  in  H, and so  x = hh1' h2b =  (hh1' h2)b, which is in   Hb.  So  Ha  is a subset of  Hb.  Similarly,  Hb  is a subset of  Ha..  So  Ha = Hb.
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4. Let G be a group with exactly pn elements, where  p  is prime.   Recall that the order of an element g is the least positive integer k such that gk = e, and that the cyclic subgroup (g) generated by g consists of the distinct elements {e, g, g2, ..., gk - 1}. Why must the order of g be a power of p?
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The order  k  of the element  g  is known to be the number of elements in the cyclic subgroup generated by  g.  By LaGrange's Theorem, the order of a subgroup divides the order of the group  G, which is   pn.  The only factors of  pare powers of p  since  p  is prime.
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5. Let  H  be a subgroup of a group  G.  On  G  define a relation  ~  on G by g ~ k  if and only if  k¢g  is in H.

(a) Prove that ~ is an equivalence relation.
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Reflexive:  g ~ g  because   g¢g = e, which is in  H.
Symmetric:  If  g ~ k  then  k¢g  is in  H.  So  (k'g)'  is in H.  But  (k'g)' = g'k, and therefore, k ~ g.
Transitive:  If  g ~ k  and  k ~ l, then  k'g  and  l'k  are in  H.  Then (l'k)(k'g) = l'g  is in  H and  g ~ l.
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(b) Prove that the equivalence classes of ~ are the left cosets of H in G.
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We show that  [a] = aH.
If  x  is in  [a], then  x ~ a and  a'x  = h  in  H.  Hence,  x = ah, which is in aH.  So,  [a]  is a subset of  aH.
If  x  is in  aH, then x = ah for some  h  in  H.  then a'x = h, and  x ~ a, so  x  is in  [a].  So,  aH  is a subset of  [a].
Therefore,  [a] = aH.
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(c) How would you define an equivalence relation on G so that the equivalence classes are the right cosets of H in G?
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g ~ k  if and only if  gk'  is in  H.
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