Algebra (647), Homework 7, solutions.
1. II.6: 1.
First we verify the identity element.
(e, e)(g, h) = (eθ(e)(g), eh) = (g , h) and (g, h)(e, e) = (gθ(h)(e), he) = (ge, he) = (g , h).
Then we check for inverses.
(g, h)(θ(h−1)(g−1), h−1) = (gθ(h)(θ(h−1)(g−1)), hh−1) = (gg−1, hh−1) = (e, e).
Similarly,
(θ(h−1)(g−1), h−1)(g, h) = (θ(h−1)(g−1)θ(h−1)(g), h−1h) = (θ(h−1)(g−1g), e) = (θ(h−1)(e), e) = (e, e).
Finally we check associativity!
(g, h)[(k, l)(m, n)] = (g , h)(kθ(l)(m), ln) = (gθ(h)(kθ(l)(m)), hln) = (gθ(h)(k)θ(hl)(m), hln).
Associating the other way
[(g, h)(k, l)](m, n) = (gθ(h)(k), hl)(m, n) = (gθ(h)(k)θ(hl)(m), hln).
2. II.6: 2.
(a)
We check
α(ai)α(aj) = asiasj =as(i+j)=α(ai+j).
So
α
is a homomorphism. If
α(ai) = e
, then
asi =e
so
p
divides
si
.
Since
s
is prime to
p
,
p
divides
i
, so
ai=e
. This shows
α
is injective,
hence it is an isomorphism since
Cp
is nite.
(b)
The map is clearly a homomorphism if it is well-dened. We need to
check that
bq
goes to the identity homomorphism.
αq(ai) = asqi=ai
so
αq= 1Cp
.
(c)
The elements of the group are all
(ai, bj)
. This element can be written
as
(ai, e)(e, bj) = aibj
, so the elements
(a, e),(e, b)
are generators.
As far as the relations, the rst two are clear. For the third we calculuate
ba = (e, b)(a, e) = (eθ(b)(a), be) = (as, b) = asb.
These three relations are sucient to prove that there are
pq
elements
(by writing all elements in the form
bjai
) so since that is the number of
elements in the semidirect product, that must exhaust the relations!
3. II.7: 4.
You have to look back on p. 100 at the denition of \nilpotent."
Suppose
γr
is eventually
(e)
. Let
m
be the smallest
m
so that
γm= (e)
.
Then
(e)6=γm−1< C(G)< G
.
This gives
G/γm−1→G/C(G)
. Now the same argument tells us that
γm−2/γm−1< C(G/γm−1)) →C(G/C (G))
so
γm−2< C2(G)
.
Now assume
γm−i< Ci(G)
. This gives
γm−i−1/γm−i< C(G/γm−i)→C(G/Ci(G)).
It follows that
γm−i−1< Ci+1
. Now since
G/γ1
is abelian, so is
G/Cm−1.
So
Cm=G
.
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