Posted: December 3rd, 2015

Introduction to Abstract Algebra

Introduction to Abstract Algebra

Homework Assignment

Attempt all the questions on this sheet

(A1) Show that cyclic groups are abelian.
(A2) In each of the following groups G, write down the cyclic subgroup generated
by g.
(a) G = S, g = exp(2pi/7).
(b) G = Z/12Z, g = 8.
(c) G = GL2(R), g = ( 0 1
-1 0 ).
(d) G = R/Z, g = 5/7.
(e) G = D4, g = ?3.
(A3) Which of the following groups G are cyclic? Justify your answer for each,
and if G is cyclic then write down a generator.
(a) G = kZ (where k is a non-zero integer).
(b) G = Z/mZ (where m is a positive integer).
(c) D3.
(A4) For the following groups G and subgroups H, write down the (left) cosets
of H in G and determine the index [G : H].
(a) G = 2Z and H = 6Z.
(b) G = U4 and H = U2.
(c) G = D4 and H = h?1i.
(d) G = R
* and H = {a ? R : a > 0}.
(A5) There are four elements of order 5 in R/Z; find them. There are eight
elements of order 3 in R
2/Z
2
; find them.
(A6) In Z
2 we let i = (1, 0) and j = (0, 1) as usual. Write 2Z
2 = {(2a, 2b) :
a, b ? Z}. Convince yourself that 2Z
2
is a subgroup of Z
2 of index 4 and
that
Z
2
/2Z
2 = {0, i,j, i + j}.
Write down an addition table for Z
2/2Z
2
.
(B1) In this exercise, you will show using contradiction that R
*
is not cyclic.
Suppose that it is cyclic and let g ? R
* be a generator. Then R
* = hgi. In
particular, |g|
1/2 ? R
* and so |g|
1/2 = g
m for some integer m. Show that
the only solutions to this equation are g = ±1. Where’s the contradiction?
(B2) Let ß ? C
* and write ß = reif where r, f ? R and r > 0. Show that
ßS = rS. What does the coset rS represent geometrically?
(B3) Let G be a group of order p, where p is a prime number. Let H be a
subgroup. Show that H must either equal G or the trivial subgroup {1}
(Hint: Use Lagrange’s Theorem). Deduce that if g ? G is not the
identity element, then G = hgi.
(B4) Let a ? [0, 1). Show that a ? R/Z has finite order if and only if a is
rational.
(C1) Show that every non-zero element of R/Q has infinite order.
(C2) Let v be a column vector in R
2 and H a subgroup of GL2(R). We define
the orbit of v under H to be the set
Orb(H, v) = {Av : A ? H}.
Observe Orb(H, 0) = {0}. Suppose v 6= 0.
(i) Let H = {?I2 : ? ? R
*}. Show that H is a subgroup of GL2(R).
Describe and sketch Orb(H, v).
(ii) Let H = SO2(R) (this is the subgroup of rotation matrices—see
Sections IV.7 and IX.4 of the lecture notes). Describe and sketch
Orb(H, v).
(iii) Finally, let H = GL2(R). With the help of (i) and (ii) explain why
Orb(H, v) = R
2{0}.
(C3) Let v be a column vector in R
2
. In the previous assignment we defined the
stabilizer of v to be
Stab(v) = {A ? GL2(R) : Av = v},
and showed that it is a subgroup of GL2(R). Now let v and w be non-zero
column vectors in R
2
.
(i) Show that there is some B ? GL2(R) such that Bv = w. Hint: Use
part (iii) of (C2).
(ii) Let U = {C ? GL2(R) : Cv = w}. With the help of (i), show that U
is a left coset of Stab(v).
(iii) Show that Stab(v) = {B-1AB : A ? Stab(w)} where B is as in (i).

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