Solutions, Algebra hw, Week 3 by Bracket 3.4. Prove that ... - Archives

3.20. If H is a subgroup of G, then by the centralizer C(H) of H we mean the set
C(H) = x ∈ G xh = hx for all h ∈ H .
Prove that C(H) is a subgroup of G.
Note that x ∈ C(H) iff hxh−1 = x for all h ∈ H. Clearly, heh−1 = hh−1 = e, so e ∈ C(H).
Suppose that a, b ∈ C(H). Then for any h ∈ H, we have hbh−1 = b. Taking the inverse, we obtain
b−1 = (hbh−1 ) −1 = hb−1h−1 . Furthermore, we have a = hah−1 , so ab−1 = (hah−1 )(hb−1h−1 ) =
h(aeb−1 )h−1 = h(ab−1 )h−1 . Thus ab−1 ∈ C(H), and it follows that C(H) is a subgroup of G from
the One-Step Subgroup Test.
Alternatively, we can see that C(H) =
h∈H C(h), and we can apply Problem 3.14 to see that this
intersection is a group.
3.27. Suppose a group contains elements a and b such that |a| = 4, |b| = 2, and a 3 b = ba. Find |ab|.
Since |a| = 4 and |b| = 2, we have a 4 = e and b 2 = e.We then multiply ba = a 3 b on the left by a
and on the right by b to obtain abab = (ab)(ab) = a 4 b 2 = ee = e. Note that ab = e, since that would
mean b is the inverse of a, and Problem 3.4 would guarantee that a and b have the same order. Thus,
(ab) 2 = e and (ab) n = e for any smaller positive integer power n, so |ab| = 2.
3.31. Let G be the symmetry group of a circle. Show that G has elements of every finite order as well as
elements of infinite order.
For any positive integer n, the element R 360/n must have order n. If we choose any irrational
number x such that 0 < x < 360, then Rx is of infinite order.
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