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Math 481, Abstract Algebra, Winter 2004
Problem Set 6, due Friday, February 13
6.31 Show that the mapping a → log10 a is an isomorphism from R+ under multiplication to R
under addition.
Let’s call the map in question φ. Now φ is obviously well defined (there is no ambiguity
about its definition), and it is not difficult to see that it is a homomorphism if we remember
a little fact from calculus 142 (that log10 (ab) = log10 (a) + log10 (b) for all a, b ∈ R+ ). So for
a, b ∈ R+ we have φ(ab) = log10 (ab) = log10 (a) + log10 (b) = φ(a) + φ(b) as required. Now if
a ∈ R then 10a ∈ R+ and φ(10a ) = log10 (10a ) = a, so φ is onto, and if log10 (a) = log10 (b),
then elog10 (a) = elog10 (b) , that is a = b, so the map is injective. We conclude that φ is an
isomorphism.
6.35 Let a belong to a group G and let |a| be finite. Let φa be the automorphism of G given by
φa (x) = axa−1 . Show that |φa | divides |a|. Exhibit an element a from a group for which
1 < |φa | < |a|.
To show that |φa | divides |a| (here |φa | denotes the order of φa as an element of Aut(G)), it
is enough to demonstrate that (φa )|a| = id where
id is the identity
automorphism (this by corollary 2 of theorem 4.1). But note that (φa )|a| (x) = (φa )|a|−1 (axa−1 ) = (φa )|a|−2 (a2 xa−2 ) =
· · · = a|a| xa−|a| = x for all x ∈ G so that (φa )|a| = id as required.
Note that in an Abelian group G, φa (x) = axa−1 = aa−1 x = x for all x ∈ G (that is, |φa | = 1).
Thus to find an example of the type required we must look to non-Abelian groups—but it can’t
be S3 , because this group only has elements of prime order. So let’s try D4 . We’ll take a to
be the rotation by 90 degrees, and thus |a| = 4. We claim that |φa | = 2. We have that
(φa )2 (x) = a2 xa−2 where a2 = a−2 is rotation by 180 degrees,and it is easy to check (by
looking at a multiplication table for D4 ) that a2 xa2 = x for all x ∈ D4 .
6.42 Prove that Q, the group of rational numbers under addition is not isomorphic to a proper
subgroup of itself.
This is not difficult to do after establishingthe fact that if φ is an isomorphism from Q under
addition to a subgroup of itself, then φ ab = ab φ(1). Indeed, with this observation we note
that if φ(1) = dc , then for all ab ∈ Q,
φ
ac−1
db−1
=
ac−1 c
a
ac−1
φ(1)
=
= .
bd−1
bd−1 d
b
Thus φ is onto Q, and thus the image of φ is not proper.
So suppose that φ is an isomorphism of Q with some subgroup of Q. Then


a
1
1
 = φ 1 + · · · + φ 1 = aφ 1
= φ
+
·
·
·
+
φ
b
b
b}
b
b
b
|
{z
|
{z
}
a times
for all
a
b
a times
∈ Q. But we can also say that
b
1
= bφ
φ(1) = φ
b
b
and thus
So we conclude that φ
1
φ(1)
φ
=
.
b
b
a
b
= ab φ(1) as required.
7.15 Suppose that |G| = pq, where p and q are prime. Prove that every proper subgroup of G is
cyclic.
By Lagrange’s theorem, any subgroup of G must have order dividing |G| = pq. The order
of any proper subgroup of G must therefore be a proper divisor of pq, either p, q, or 1. A
subgroup of order 1 is obviously cyclic—for subgroups of order p or q, we resort to corollary 3
of Lagrange’s theorem (that any group of prime order is cyclic).
7.21 Suppose that G is an Abelian group with an odd number of elements. Show that the product
of all of the elements of G is the identity.
By Lagrange’s theorem, no element of G can have order two (two doesn’t divide an odd number)
so no element is its own inverse. Inverses are also unique and thus writing |G| = 2n + 1 we can
choose distinct elements g1 , . . . , gn ∈ G such that G = {e, g1 , g1 −1 , . . . , gn , gn −1 } (the implication being that {e, g1 , g1 −1 , . . . , gn , gn −1 } are the distinct elements of G). Now it is clear that
the product of all the elements of G gives the identity, that is, that eg1 g1 −1 g2 g2 −1 · · · gn gn −1 =
e.
7.34 Prove that a group of order 12 must have an element of order 2.
Let G be a group of order 12. By Lagrange’s theorem, an element in G can have order 1, 2, 3,
4, 6, or 12. Note that only the identity has order 1. Now, if G contains an element g of order
6
12 (that is, if G is cyclic),
if
n then |g | = 2 (we use the fact—shown many times in class—that
|a| = n and k|n, then a k = k). Similarly, if G contains an element g of order 6, then |g 3 | = 2,
and if G contains an element g of order 4, then |g 2 | = 2. So we conclude that G contains an
element of order 2 unless all element of G (except the identity) have order 3. This gives us
exactly 11 elements of order 3. But this cannot occur, because elements of order 3 in a group
occur in pairs. That is, for each g ∈ G\{e}, we have g −1 ∈ G\{e}, and g −1 is distinct and
unique (but we cannot pair off eleven elements). We conclude that G must have an element
of order 2 as required.
7.46b Calculate the order of the group of rotations of a regular octahedron (a solid with eight
congruent equilateral triangles as faces).
We will use the fact that if G is a finite group of permutations on a set S, then |G| =
|stabG (i)||orbG (i)| for any i ∈ S. For us, S consists of the faces of the regular octahedron, and
choosing one of the faces to be the “top” face, call it i, we see that |stabg (i)| = 3 (the only
rotations which fix the top face are rotations through that face, and there are three rotations of
an equilateral triangle). On the other hand, |orbG (i)| = 8 (clearly we can reorient the regular
octahedron so that our original face lies where any of the other faces originally abode), so that
|G| = 24.