Math 3124 Monday, October 6
Fourth Homework Solutions
1. We have
10 ∗0+9∗6+8∗6+7∗9+6∗0+5∗3+4∗9+3∗2+2∗5+1∗6=7
mod 11, where as the result should be 0 mod 11. Therefore we want to interchange
two digits so that [4] is added to the result. If we interchange adjacent numbers [a],
[b], the result is that [b]−[a]is added, so we want to find a,bso that [b]−[a] = [4]. A
glance at the above numbers tells us that the only possibility is interchanging the 9 and
2: this adds [2]−[9] = [4]to the result. Therefore the correct ISBN is 0-669-03295-6.
2. Clearly the order is at least 2 (only the identity has order 1). But
[1] [2]
[0] [4][1] [2]
[0] [4]=[1] [0]
[0] [1],
so the order is exactly 2.
3. Let g∈G. If g=e, then obviously g2=e. On the other hand if g6=e, then ghas order
2 and again we deduce that g2=e. Let a,b∈G. We need to prove ab =ba. Using
the above with g=a,b,ab, we obtain
a2=b2=abab =e.
Multiplying the last equation on the left and right by aand brespectively, we get
a2bab2=ab. Since a2=b2=e, we conclude that ebae =ab and hence ba =ab for
all a,b∈G, which proves that Gis abelian.
4. Problem 14.31 on page 81.
Prove that if Gis a group and a,b∈G, then o(a−1ba)= o(b).
From the September 26 ungraded homework, we have (a−1ba)n=a−1bna. Also
a−1bna=eif and only if bn=e. Therefore for n∈N,
(a−1ba)n=e⇐⇒ a−1bna=e⇐⇒ bn=e.
Thus the powers of a−1ba which are eare exactly the same as the powers of bwhich
are e. Since o(g)is the least positive integer msuch that gm=e, it follows that
o(a−1ba) = o(a), as required.
Remark. If one assumes that o(b) = n, then it is not sufficient to prove that (a−1ba)n=
e, because this only shows that o(a−1ba)≤n(or divides n).