Some Model Answers for HW on Chapter 2
#12: For any integer n>2, show that there are at least two elements in U(n)
that satisfy x2=1.
Clearly 1 is one such element. The other is n-1. Every U(n)
contains n-1, since n-1 is always relatively prime to n. Furthermore
(n-1)2 = n2 -2n+1 = 1 mod n
#16: Prove that (ab)-1 = b-1 a-1
Notice that (ab)(b-1 a-1 )=a(b b-1)a-1 = a a-1 = e
#20: For the same reason as in #16
(a1 a2 a3 . . . an) (an-1 an-1-1 . . . a1-1)=e
#26: Prove that if (ab)2 =a2b2 in a group, then ab=ba.
Write out the powers: If abab=aabb, then we can cancel a on the left
and b on the right, and get
ba=ab
#29: Let G be a finite group. Show that the number of elements x of G such
that x3=e is odd.
Let H denote the set of all elements with this property.
Certainly, e is in H.
Next, suppose x is in H. Then x-1 is also in H, because x3=e implies
that (x-1)3=e (just multiply both sides with x-1 three times).
Note also that x is not equal to x-1, because if x was equal to x-1, then x2=e.
Since also x3=e, it would follow that x=e.
This means that each element in H (besides e) can be paired up with its
unique inverse, giving an even number of elements in H. Also
counting e, it makes an odd number of elements in H.
Show that the number of elements in x in G such that x2 ≠ e is even.
Again let H denote the set of all elements with this property. e is not
in H this time. But if x is in H, then x-1 is also in H, because if (x1 2
) =e, so would be x2=e. Note again that if x is in H, x-1 can’t equal its
inverse, because otherwise x2=e. So all elements in H can be ordered
into pairs, giving an even number in H.
#33. Prove that if G is a group with the property that the square of every
element is the identity, then G is abelian.
Let a and b be in G. Since the square of every element is e, we have
a2=e
b2=e
(ab)2=e. (again, because every element’s square is e, including the
element ab.
Hence (ab)2=a2b2 for all a and b in G. By the previous problem #26
this shows that G is abelian.
#35. Prove that the set H={1,2,. . . , n-1} is a group under multiplication
modulo n if and only if n is prime
►
assume that H is a group, but, by way of contradiction, that n is not
prime. Then n can be factored: n=pq, with 1<p,q<n-1. But this means
that p*q=0 mod n. But 0 is not in H, so the group is not closed (or
more precisely, the binary operation is not well defined on H).
contradiction: n must be prime.
◄
assume that n is prime. Then by exercise 13 in Chapter 0, ax=1 mod n
has a solution for every a in H. This means every a in H has an
inverse.
The identity is in H, and multiplication is associative, so H is a group.