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

3.4. Prove that in any group, an element and its inverse have the same order.
Solutions, Algebra hw, Week 3
by Bracket
Let g be an element in some group G. Suppose that g has finite order n, so g n = e. We multiply
on the right by (g −1 ) n , to obtain g n (g −1 ) n = (g −1 ) n . Using the definition of the identity element and
and the inverse of an element, we can collapse the left side until we are left with e = (g −1 ) n , so g −1
has order n as well.
Alternatively, suppose that g is of infinite order. Then g −1 must be of infinite order as well. To see
this, we merely need note that g is the inverse of g −1 , and thus if g −1 has finite order then g must
have finite order from the first part of the proof.
3.6. Let x belong to a group. If x 2 = e and x 6 = e, prove that x 4 = e and x 5 = e. What can we say about
the order of x?
Clearly x = e, since otherwise x n = e for all n ∈ . If x 4 = e, then x 6 = (x 2 )(x 4 ) = x 2 e = x 2 = e,
which is a contradiction. Similarly, if x 5 = e, then x 6 = x(x 5 ) = xe = x = e. We cannot rule out the
possiblity that x 3 = e, so x must have order of either 3 or 6.
3.10. Prove that an Abelian group with two elements of order 2 must have a subgroup of order 4.
Let G be an Abelian group with two distinct elements a and b such that a 2 = b 2 = e, and form the
set
G ′ = e, a, b, ab .
It is straightforward to verify that G ′ is a subgroup of G, and that e, a, b, and ab are distinct so
that |G ′ | = 4.
3.12. Suppose that H is a proper subgroup of under addition and H contains 18, 30, and 40. Determine
H.
Since H is closed under addition, H must contain every linear combination of 18, 30, and 40. Note
that gcd(18, 30, 40) = 2 so that we can write 2 = r · 18 + s · 30 + t · 40 for some integers r, s, and t,
and thus 2 ∈ H. Since H = we must have H = 2.
3.14. If H and K are subgroups of G, show that H ∩ K is a subgroup of G. (Can you see that the same
proof shows that the intersection of any number of subgroups of G, finite or infinite, is again a subgroup
of G?)
We can see that H ∩ K is nonempty since both H and K must contain e to be groups. Suppose
that a, b ∈ H ∩ K. Since b ∈ H and b ∈ K and H and K are both groups, we must have b −1 ∈ H
and b −1 ∈ K. Furthermore, since a ∈ H and a ∈ K and both H and K are closed under the group
operation, we have ab −1 ∈ H and ab −1 ∈ K. It follows that ab −1 ∈ H ∩ K, so H ∩ K is a subgroup
by the One-Step Subgroup Test. This proof generalizes to the intersection of an arbitary family G of
subgroups of G, since if a and b are in every element of G then a −1 , b −1 , and ab must be in every
element of G since every element of G is a group.
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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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