Math 3124 Monday, November 17
Ninth Homework Solutions
1. Problem 22.6 on page 114. Prove that every quotient group of a cyclic group is cyclic.
Let Gbe the cyclic group, and suppose the quotient group is G/Hwhere HG. Since
Gis cyclic, we may write G=hxifor some g∈G. Then every element of Gis of the
form xnfor some n∈Z. Since the general element of G/His of the form Hg for
some g∈G, we see that every element of G/His of the form Hxnfor some n∈Z.
But Hxn= (H x)nand we conclude that G/H=hHxi. Therefore G/His cyclic and is
generated by Hx.
2. (a) θ(AB) = det(AB) = det(A)det(B) = θ(A)θ(B), so θis a homomorphism. It
is onto because given x∈Z#
5, we have θx0
0 1=x, and the foregoing matrix is
in GL2(Z5).
(b) A∈kerθif and only if det(A)=[1].
(c) Since K=kerθand θis onto (from above), this follows immediately from the
Fundamental Homomorphism Theorem.
(d) Since the property of being cyclic is invariant under isomorphism, it will be suffi-
cient to determine whether Z#
5is cyclic. The answer is yes, because Z#
5=h[2]i.
3. Define θ:G→Hby θ(x) = x4. Since Z#
509 is abelian, we have θ(xy)=(xy)4=
x4y4=θ(x)θ(y)and so θis a homomorphism. It is onto by the definition of H.
Therefore by the Fundamental Homomorphism Theorem G/kerθ∼
=H. We need
to determine kerθ. Now [208]∈ker θbecause [208]4= [−1]2= [1]. Also [208]2=
[−1]6= [1], consequently o([208]) = 4. Therefore 4
|kerθ|. Since |G|=508 =4∗127
and 127 is prime, we see from Lagrange’s theorem that |kerθ|=4 or 508. But the
latter is not possible because that would imply kerθ=G, which is clearly not the
case. Therefore |ker θ|=4 and we deduce that |H|=127.
4. Problem 24.17 on page 125. Prove that (a+b)2=a2+2ab +b2if and only if Ris
commutative.
Since (a+b)2=a2+ab +ba +b2, we have (a+b)2=a2+2ab +b2if and only if
ab +ba =2ab, that is if and only if ba =ab. This last equality is exactly the condition
for Rto be commutative.