# Download Algebra 411 Homework 5: hints and solutions

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```Algebra 411 Homework 5: hints and solutions
Stephen Coughlan
As usual, these are only hints and not intended to be a substitute for full
solutions. There is often more than one valid method. Also, there may be mistakes!
6.1 We had a theorem in class which stated that the order of an element
(g1 , . . . , gn ) in a direct product G1 × · · · × Gn is given by the formula
o((g1 , . . . , gn )) = lcm(o(g1 ), . . . , o(gn )).
Using this formula and our knowledge of cyclic groups, we get
(a) o(4) =
18
hcf(4,18)
= 9 and o(9) = 2, so o((4, 9)) = lcm(9, 2) = 18.
(b) o(7) =
12
hcf(12,7)
= 12 and o(5) =
8
hcf(5,8)
= 8 so o(7, 5) = lcm(12, 8) = 24.
(c) o(8) = 9, o(6) = 3, o(4) = 2 so o((8, 6, 4)) = lcm(9, 3, 2) = 18.
(d) o(8) = 9, o(6) = 17, o(4) = 5 so o((8, 6, 4)) = lcm(9, 17, 5) = 5·9·17 = 765.
6.2 The group Zm × Zn is cyclic if m and n are coprime, i.e. hcf(m, n) = 1.
(What is the condition if we have the product of three or more cyclic groups?
There are several ways to answer this.)
(a) hcf(12, 9) = 3 so not cyclic.
(b) hcf(10, 85) = 5 so not cyclic.
(c) hcf(4, 6) = 2 so not cyclic.
(d) hcf(22, 21) = hcf(22, 65) = hcf(21, 65) = 1 so is cyclic.
6.6 We show that G × H is abelian iff G and H are abelian. The product of
more than two groups is the same. We also discussed this a little bit in class.
Suppose G and H are abelian, and take elements (g1 , h1 ), (g2 , h2 ) of G × H.
We check
(g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ) = (g2 g1 , h2 h1 ) = (g2 , h2 )(g1 , h1 ).
The middle equality is because G and H are abelian groups. Thus G × H is
abelian. To prove G × H is abelian implies that G and H are abelian, just write
the same argument in the reverse order!
6.10 Write G = Z2 × Z4 . There are several cyclic subgroups of G, each given
by choosing a single element g in G, and taking the cyclic group hgi generated
by that element. We list these:
h(0, 0)i ∼
= {0}, h(1, 0)i ∼
= Z2 , h(0, 1)i = h(0, 3)i ∼
= Z4 ,
1
h(0, 2)i ∼
= Z2 , h(1, 1)i = h(1, 3)i ∼
= Z4 , h(1, 2)i ∼
= Z2
This is eight elements in G, but two pairs of them give the same cyclic
group, so we only get six different subgroups. I have also stated above what
each subgroup is isomorphic to, although this was not part of the question.
There is also the group G itself, which is a subgroup. Finally, we must
consider groups generated by two or more elements. The only possible new
group that we get here is
h(1, 0), (0, 2)i = {(0, 0), (1, 0), (0, 2), (1, 2)} ∼
= Z2 × Z2 .
Any other choice of two elements would give one of the groups already listed.
(Why?)
There is no way to get a new group by choosing more than two generators.
(You can check this yourself!)
In total, there are 6 + 1 + 1 = 8 subgroups of G.
Extra 1 One such isomorphism f : Z6 → Z2 × Z3 is given by the map sending
0 7→ (0, 0), 1 7→ (1, 1), 2 7→ (0, 2), 3 7→ (1, 0), 4 7→ (0, 1), 5 7→ (1, 2).
Extra 2 This problem is actually solved in chapter 12 of the textbook. We
have to check that f −1 is a bijection, and that f −1 (xy) = f −1 (x)f −1 (y). We
check only the second of these conditions. The first is easier, and is left to you.
Since f is a bijection, every element of H can be written in the form f (a)
for some a in G. In other words, given x, y in H, we can think of x = f (a) and
y = f (b) when convenient. Now, we want to check that
f −1 (xy) = f −1 (x)f −1 (y),
but this can be written in terms of a and b:
f −1 (f (a)f (b)) = f −1 (f (a))f −1 (f (b)) = ab,
where the last equality is because f −1 is the inverse of f . Then on the left hand
side, we note that f (a)f (b) = f (ab) because f is an isomorphism. Hence we
need to check that
f −1 (f (ab)) = ab.
Why is this true?
Extra 3 Here are the only two isomorphisms from Z4 to Z4 :
0 7→ 0, 1 7→ 1, 2 7→ 2, 3 7→ 3,
and
0 7→ 0, 1 7→ 3, 2 7→ 2, 3 7→ 1.
The first of these is also called the identity map. The second of these is actually
multiplication by 3 modulo 4!
2
There are no other isomorphisms from Z4 to Z4 . In fact, because f (x + y) =
f (x) + f (y), any isomorphism is completely determined by where we send 1 to.
In other words, if we choose f (1) = 1, then since f (1 + 1) = f (1) + f (1), it
follows that f (2) = 2, and likewise f (3) = 3, f (0) = 0. Similarly if we had
started from f (1) = 3, we would get the second isomorphism. What goes wrong
if we try f (1) = 2?
By the same arguments, isomorphisms from Z8 to Z8 are determined by the
value of f (1). Here are the possibilities:
0 7→ 0, 1 7→ 1, 2 7→ 2, 3 7→ 3, 4 7→ 4, 5 7→ 5, 6 7→ 6, 7 7→ 7,
0 7→ 0, 1 7→ 3, 2 7→ 6, 3 7→ 1, 4 7→ 4, 5 7→ 7, 6 7→ 2, 7 7→ 5,
0 7→ 0, 1 7→ 5, 2 7→ 2, 3 7→ 7, 4 7→ 4, 5 7→ 1, 6 7→ 6, 7 7→ 3,
0 7→ 0, 1 7→ 7, 2 7→ 6, 3 7→ 5, 4 7→ 4, 5 7→ 3, 6 7→ 2, 7 7→ 1.
Can you see any patterns?
Extra 4
(a) The map f (x) = x−1 is one-to-one because every element of G has a
unique inverse, i.e. two different elements of G can not have the same
inverse. The map is onto because any element x in G is the inverse of
x−1 .
(b) Be careful with brackets here:
f (xy) = (xy)−1 = y −1 x−1 , f (x) = x−1 , f (y) = y −1 , f (x)f (y) = x−1 y −1 .
(c) We want to check that f (xy) = f (x)f (y). By part (b), this is equivalent
to checking y −1 x−1 = x−1 y −1 . This is only true G is abelian. It may not
be true for a general group!
3
```