Mathematics 581   Problem Set 5   Due Monday, May 13

1. Let  G  be any group.  Let  f  be the function from  G  into  G  defined by f(a) = a'  for every  a  in  G.

(a)  Show that  f is  one-to-one and onto.
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one-to-one:  if  f(a) =  f(b), then  a' = b' and then  a = (a')' = (b')' = b.
onto:  for any  a  in  G,   f(a') = a.
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(b)  Let  H  be any subgroup of  G.  Show that  f  maps H, which is a subset of  G, onto  H.
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If  a  is in  H, then f(a) = a' is in  H.  So  f(H) is a subset of  H.  On the other hand, if  b is in  H, then   b'  is in  H  also.  But f(b') = (b')' = b.  So  b  is in  f(H).  So,  H  is a subset of  f(H).
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(c)  Let  a  be any element of  G.  Show that  f  maps the left coset  aH, which is a subset of  G, onto the right coset  Ha¢.
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If  x  in  in  aH, then  x = ah  for some  h  in  H.  Then f(x) = x' = h'a', which is in  Ha'.  So, f(aH)  is a subset of  Ha'.  On the other hand, if  x  is in  Ha', then  x = ha' for some  h  in  H.  Then  x' = ah', which is in  aH.  So  f(x') = x, and  Ha'  is a subset of  f(aH).
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2.  Let  f be an isomorphism of a group  G  onto a group   C.  Let  a  be an element of  G.  Show that the order of the element  f(a)  in  C  is equal to the order of the element  a. (Two cases:  order  a  is infinite, and order  a  is a finite positive integer  n.)
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Exploit the fact that an isomorphism is one-to-one.
If the order of  a  is infinite, then all of the elements  ak  are distinct.  Then all of the elements f(ak) = f(a)k must also be distinct,  which is sufficient for the element f(a)  to have infinite order.
On the other hand, suppose that the element  a  has finite order  n.  Then  an = e  and all of the elements  e, a, a2, a3, ... an - 1  are distinct.  Therefore,  f(a)n = f(an) = f(e) = the identity element in the group  C, and because  f is one-to-one, the elements  f(e), f(a),f(a)2,  ...,  f(a)n - 1 are all distinct.  This guarantees that the order of f(a)  is  n, also.
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3.  This is a generalization of problem  2.  Let  be a homomorphism of a group  G  into a group   C.  Let  a  be an element of  G  and let the order of  a  equal n,  finite.  Show that the order of  f(a)  is finite and a factor of  n.
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We know from problem 3 that f(a)n = f(e) = the identity element in the group  C.  Let  k  be the smallest positive integer such that  f(a)k = the identity element in the group  C.  Then  n = kq + r, where  r  is  0  or greater but less that  k.  Then  f(a)r= f(a)n - kq= f(a)n (f(a)k)-q = the identity element in the group  C.  therefore,  r = 0, and  k  is a factor of  n.
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4. (a)  Show that if  s   and  t  are disjoint cycles then  st =  ts.
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The set of elements moved by the cycle s  is disjoint from the set of elements moved by the cycle  t.  Therefore, it doesn't matter which cycle goes first.  The end result is the same.
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(b)  Show that the order of the permutation  (1 2 3)(4 5) is 6.
***Let  s = (1 2 3)  and  t = (4 5).  Then, because   st =  ts,  ((1 2 3)(4 5))6 = (st)6 = s6t6  = (1).  If  k  is any positive integer  for which  (st)k = sktk  = (1), then  sk  and  tk must both equal  (1), because  the sets of elements moved by  and  t  are completely disjoint.  Because s = (1 2 3),  k  must be a multiple of  3,  and because  t = (4 5),  k  must be a multiple of  2.   Hence,the least common multiple of  2  and  3  divides  k, that is,  6  divides  k.  Therefore, the least positive integer  k  for which  (st)k  = (1) is  k = 6, and  6  must be the order of  the element  st.
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(c)  Show that the order of  (1 2 3 4)(5 6 7)  is  12.
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Same argument as part (b).
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(d)  What is the order of  (1 2)(3 4 5)(6 7 8 9)? (It is not 24.)
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The order is  12.  Same argument as part (b).  The least common multiple of  2, 3, and  4  is  12.
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(e)  What is the order of  (1 2 3 4 5 6)(11 12 13 14 15 16 17 18 19 20)?
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The order is  30.   Same argument as part (b).  The lease common multiple of  6  and  10  is  30.
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(f)  Prove that if a permutation s in Sn  can be written as a product of disjoint cycles s and the cycles have lengths   m1, m2, ..., mk, then the order of  s is the least common multiple of  m1, m2, ..., mk.
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Same argument as part (b).
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5.  Put  a Cartesian coordinate system onto the plane.  Let F  be the geometric figure in the plane which consists only of the points along the x-axis with  integer coefficients.  That is, F =  { ..., (-2,0), (-1, 0), (0, 0), (1, 0), (2, 0), ...}. These points are spaced 1 unit apart.  Remember that a symmetry of  F is nothing more than a one-to-one map of  onto  F  which preserves distances.   Find all of the symmetries of F.  (Hint:  think of  F  as like a regular polygon with infinitely many sides extending to the left and right of any starting point.  Think of what might be analogues for  F  of rotations and reflections  on  an  n-gon.)
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The analog  r  of a clockwise rotation through one vertex is just a shift one unit to the left.
r((n, 0)) = (n - 1, 0).   This preserves distances.  The analogue of the reflection through vertex through the first vertex on the polygon is  the reflection l  where  l((n, o) = (-n, 0).  It just interchanges the positive and negative halves of the number line, leaving  the origin fixed.

The subgroup  H  that fixes the origin is  H = { (1) , l }.  The elements in the symmetry group are then
..., r-2, r-1, e, r, r2, ...  and  ..., r-2l  , r-1l  , l  , rl  , r2l  , ...  . You can check that   lr = r-1l..
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