Mathematics 581  Problem Set 4  Due Monday, April 29

1. Recall that the index of a subgroup H in a group G is the number of left cosets of H in G. Define a function f of the set of left cosets into the set of right cosets by

f (aH) = Ha¢ .

Prove

(a) f is well-defined,
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Let  aH = bH.  Then b'a  is in H.  Then (b'a)' = a'b  is in  H.  But  a'b = a'(b')', so that  Ha' = Hb'.
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(b) f is one-to-one,
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Let Ha' = Hb'.  Then  a'(b')'  is in H.  But  a'(b')' = a'b.  So (a'b)'  is in  H.  But  (a'b)' = b'a.  So  b'a   is in H, and aH = bH.
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(c) f is onto.
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For any  x  in  G, f(x'H) = H(x')' = Hx.
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(This is without regard to whether G is finite or whether the number of left or right cosets is finite.)
 

2. (a) If s is an element of Sn and s = (a b)(b c) with a, b, c distinct, show that s can be written as a single 3-cycle.
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(a b)(a c) = (a b c).  Just work it out.
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(b) If s is an element of Snand s = (a b)(c d) with a, b, c, d distinct, show that s is a product of two 3-cycles by inserting e = (b c)(b c) between (a b) and (c d).
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(a b)(c d) = [(a b)(b c)][(b c)(c d)] = (a  c  b)(b  d  c)
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(c) Prove that every even permutation in Sn can be written as a product of 3-cycles. (Each individual 3-cycle is even, so whether there is an even number of 3-cycles in the product or an odd number is irrelevant.)
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Every even permutation is a product of an even number of transpositions, and each successive pair of transpositions can be replaced by  a product of 3-cycles.
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3. Show that if H is a subgroup of G and the index (G: H) = 2, then H is a normal subgroup of G. (Hint. As a start you could show that for every element a not in G, aH = Ha, because each must be the set difference G - H.)
***Because there are only two left cosets and one of them is  H, the other must be  G - H.  Similarly there is one right coset other than  H, and it must also be  G - H.  But the other left coset is  aH  for some a not in  H.  Then the right coset  Ha is not H because a  is not in H.  Therefore aH = G - H = Ha.

Finally, Let g be any element of  G.  Then  g  is either in  H  or in  aH = Ha.

If  g  is in  H, then  so is  g'.  Then because  H  is closed under multiplication,  g'Hg  is a subset of  H.  On the other hand, if  h  is in  H, then  ghg'  is in  H  and then  h = g'(ghg')g is in  g'Hg.  So each of  H  and  g'Hg is a subset of the other, and they must be the same subset of G.

Now let  g  be in  aH , so that  g = ak for some  k  in H.  If  x  is in  g'Hg, then  there is an  h  in  H  so that  x = g'hg  = k'a'hak = k'(a'ha)k.
But  Ha = aH, so that  ha = ah1  for some  h1  in  H.  Then
x = k'(a'ha)k = k'(a'ah1)k = k'h1k, which is in H.  So, g'Hg  is a subset of  H.  Finally if  h  is in H, then  h = g'(ghg')g = g'(akhk'a')g.  But akhk' = a(khk')  is in  aH = Ha.  So there is an element  h2  in  H  such that  akhk' = h2a.  Then h = g'(akhk'a')g = g'(h2aa')g =  g'h2g, which is in g'Hg.  So  H  is a subset of  g'Hg.  Hence, H = g'Hg.

So for all g  in  G,  H = g'Hg, and  H  is a normal subgroup of  G.
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4. Let G and C be groups and let F be a homomorphism of G into C. Let a be any element of G and let k be any integer. Let e be the identity in G and f be the identity in C. We proved in class that F(e) = f. Since a0 = e and F(a)0 = f, we now know that F(a0) = F(a)0.

(a) Prove by induction that if k is positive then F(ak) = F(a)k.
***If  k = 1, then  F(a1) = F(a) = F(a)1.  If  F(ak) = F(a)k then
F(ak + 1) = F(aka) = F(ak)F(a) =  F(a)kF(a)  =  F(a)k + 1.
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(b) We proved in class that F(a¢ ) = F(a)¢ . Put this together with part (a) to deduce that if k is negative, then F(ak) = F(a)k.
***If  k  is negative, then  k = -m for some positive m.  Then
F(ak) = F(a-m) = F((a')m) =  F(a')= [F(a)']m. = F(a)-m = F(a)k
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[You've now shown formally that for any integer k, F(ak) = F(a)k.]
 
 
 
 
 
 
 

5. Let G = (g) be a cyclic group. Our main unresolved question about cyclic groups is to determine which elements gk other than g itself can be generators of G. Another way to phrase this is to observe that for every integer k (positive or negative) we get a subgroup (gk) of G.

Then the question becomes  for which  values of  k  is  (gk) = G?

(a) Show that if G is infinite then (gk) = G if and only if k = 1(trivial) or k = -1.
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If  k = -1, then  g  is the inverse of g-1 and  g  is in  (g-1).  So, all powers of  g  are in  (g-1).  So  (g-1) = G.
On the other hand, if  (gk) = G  for some integer  k, then  g is a power of  gk.  So, g1 = g = (gk)m = gkm for some integer  m.  But in an infinite cyclic group (g), different powers of the generator  g  cannot be the same group element.  Therefore,  1 = km.  But the only integers  k  which divide  1  are  k = +1  and  k = -1.
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(b) Show that if G is finite and has n elements, then (gk) = G if and only if the greatest common factor (n, k) = 1. (You may use everything that we did in Math 580 with integers, factors, and greatest common factors.)
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   If  (gk) = G  then   g  is an element of  (gk).  Then  g1 = g = (gk)m = gkm  for some integer  m.  But we know that in a finite cyclic group of order  n,  gn = e  and   gi = gj if and only if  i - j  is a multiple of  n.  Hence  1 - km = ns  for some integer  s.  So,
1 = ns + km, and  k  and  n are relatively prime.
Conversely if  k  and  n  are relatively prime then  1 = ns + km  for some  s  and  m.  then   g = g1 = (gn)s(gk)m =  (es)(gk)m = (gk)m.  Therefore,  g  is an element of  (gk), and  gk  generates  G.
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