Download Math 481, Abstract Algebra, Winter 2004

Math 481, Abstract Algebra, Winter 2004
Problem Set 2, Solutions
1. Prove that no column in the multiplication table of a group can have a repeat entry.
Let G be a group and suppose that some column in the multiplication table of G has a repeat
entry. Then there are a, b, c, d ∈ G such that a 6= b and ac = d = ab. So ac = bc, but by
cancellation this implies that a = b, a contradiction. We conclude that no column can have a
repeat entry.
2.17 Prove that a group G is Abelian if and only if (ab)−1 = a−1 b−1 for all a and b in G.
Suppose first that G is an Abelian group. We must demonstrate that (ab)−1 = a−1 b−1 for all
a and b in G. Of course, given a, b ∈ G, we have that (ab)−1 = b−1 a−1 (by problem 16), and
b−1 a−1 = a−1 b−1 by the Abelian property, so (ab)−1 = a−1 b−1 holds as required.
Suppose, on the other hand, that for all a, b ∈ G, (ab)−1 = a−1 b−1 . We must demonstrate
that ab = ba for all a, b ∈ G. But note that a−1 b−1 = (ab)−1 = b−1 a−1 (again by problem
16), so a−1 b−1 = b−1 a−1 for all a, b ∈ G. Multiplying both sides of the equation with ab on
the left and ba on the right we then obtain aba−1 b−1 ba = aba−1 ea−1 = ab on the left and
abb−1 a−1 ba = aea−1 ba = ba on the right, that is, that ab = ba as required.
2.31 Suppose that G is a group with the property that for every choice of elements in H, axb = cxd
implies ab = ce. Prove that G is Abelian.
We need to show that ab = ba for all a, b ∈ G. Given a, b it is enough to find x such that
axb = bxa, because then by hypothesis we may conclude that ab = ba. To finish, we simply
note that x = b−1 ab−1 works. (Coincidently, a−1 ba−1 works as well).
2.33 Prove that if G is a group with the property that the square of every element is the identity,
the G is Abelian.
It is enough to show that ab = ba for all a, b ∈ G. Note that (ab)(ba) = abba = aea = aa = e
(because squares are the identity), so that (ba) is the inverse of (ab). By hypothesis, (ab) is
the inverse of (ab). Thus because inverses are unique it must be that ab = ba as required.