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Algebra Homework #4 Solutions
3.11 Claim: Let G be a group. Then the set Aut(G)
of group automorphisms of G forms a group
under composition.
Proof. We need to verify the group axioms for the set Aut(G) under the operation
of composition.
First, we show that Aut(G) is closed under composition. We’ll need the
following:
Lemma: Let : G → G be maps. Then
i) if and
are injective then so is ,
ii) if and
are surjective then so is ,
iii) if and
are bijective then so is
,
iv) if and
are group homomorphisms then so is ,
v) if and
are group isomorphisms then so is .
Proof. To i), let x, y ∈ G, then
where the second and third implications follow if
and are
injective, respectively. Thus is
injective.
To ii), let x ∈ G, then since
is surjective, there exists x' ∈ G such that
. Since is
surjective, there exists x'' ∈ G such that .
But then
so we see that is
surjective.
To iii), combine i ) and ii).
To iv), let x, y ∈ G, then
if both and
are homomorphisms. So we indeed see that
is a homomorphism.
To v), combine iii ) and iv).
Thus we see that for automorphisms
, ∈ Aut(G) the composition
∈ Aut(G) is again an
automorphism, so Aut(G) is closed under composition.
Next we quickly verify that composition is associative. For
and for x ∈ G we
have
so that indeed , so
composition is associative.
Next, we find an identity. Let id : G → G be the identity function, which is
clearly an automorphism.
For ∈ Aut(G) and for x ∈ G note that
so that indeed
and . Thus id ∈ Aut(G) is indeed an identity.
Finally, we check that inverses exist, but we already did this in exercise 3.5.
For an isomorphism
: G → G, we previously showed that the
inverse function is again an isomorphism,
and by definition satisfies and
, so is
an inverse of
for composition. So
indeed, Aut(G) has inverses. We’ve finished showing that Aut(G) is a group under
composition.
3.14 Determining some automorphism groups.
a) We’re already show that Aut(Z) = {±id} in exercise 4.4.
b) Since Z/10Z is a cyclic group generated by 1, any homomorphism
: Z/10Z → Z/10Z
is completely defined by the image of 1. Now we also know by exercise 3.6a that
if is an
isomorphism, then it preserves orders of elements , i.e.
for all x ∈ Z/10Z. In
particular, a generator must be sent to a generator. Now in exercise ∈.16b, we
already know
that the only elements in Z/10Z that generate are 1, 3, 7, 9. It’s also easy to
see that each of
the four choices of where to send 1 gives an automorphism of Z/10Z, so we’ll
label them
accordingly:
Note that
. Now we compute the group structure on Aut(Z/10Z).
For example, for
x ∈ Z/10Z, we have
so we find that
. Continuing like this we can calculate the
multiplication
table for Aut(Z/10Z):
Notice that we have a nice group isomorphism
We also see that both
∈ Aut(Z/10Z) have order 4, i.e. they each generate. This shows
that Aut(Z/10Z) is cyclic, and we can construct two different isomorphisms
neither of which seems particularly appealing, but just
illustrates the two ways we can force
ourselves to think of Aut(Z/10Z) as a cyclic group of order 4.
c) Writing
, we see that the symmetric group
is generated by elements s, t or orders 2, 3, respectively, subject to a further
relation. Any
automorphism
is determined by the images of s, t, and as before, must preserve
the orders of elements . Now
has three elements s, st, st^{2} of
order 2, and two elements t, t^{2}
of order 3. So any automorphism must take s to one of s, st, s^{2} and t
to one of t, t^{2}. There
are only six conceivable ways of doing this:
One now checks that each of these in fact does give an
automorphism of
. Thus Aut()
just consists of these six elements. We would further like to know the structure
of Aut().
One way to do this is to know that there are only two isomorphism classes of
groups of
order six , namely cyclic of order six and
. We then just need to check if two of these
automorphisms don’t commute. In fact Another
way to see this is to note
that the center Z() is trivial, so that
conjugation by each element of
gives a different
automorphism, since there are already six of these, these fill up all of Aut().
Thus we have
the nice isomorphism
in the notation from lab.
d) The analysis of Aut(Z/8Z) follows exactly the same way as for Aut(Z/10Z) in
part b). In
the end, we find that Aut(Z/8Z) = and we
have the nice isomorphism
Incidentally, we check that each element of Aut(Z/8Z) has
order two, so that Aut(Z/8Z)
Z/∈Z × Z/∈Z.
e) Is the automorphism group of a cyclic group necessarily cyclic? Well, no, see
part d).
f) Is the automorphism group of an abelian group necessarily abelian? Well, no
either. Take for
example the abelian group Each permutation of the entries
gives a
group automorphism, and as we know, permutations of three objects don’t usually
commute.
In particular, we see that Aut() has a subgroup isomorphic to
the
permutation group
. Do you think that is the whole automorphism
group?
4.8 Subgroups of groups.
a) The subgroups of are:
and {e}, {e, t, t^{2}},
are normal subgroups.
b) The subgroups of the quaternion group Q = {±1,±i,±j,±k} where i^{2} =
j^{2} = k^{2} = −1
and ij = k, jk = i, and ki = j, are:
and every subgroup is normal.
4.9b Claim: Let
: G → G' and
: G' → G'' be
homomorphisms of groups. Then
Proof. Obvious.
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