1. Recall that the index of a subgroup H in a group G is the number of left cosets of H in G. Define a function f of the set of left cosets into the set of right cosets by

f (aH) = Ha¢ .

Prove

(a) f
is well-defined,

***

Let aH = bH. Then b'a
is in H. Then (b'a)' = a'b is in H. But a'b
= a'(b')', so that Ha' = Hb'.

***

(b) f
is one-to-one,

***

Let Ha' = Hb'. Then a'(b')'
is in H. But a'(b')' = a'b. So (a'b)' is in
H. But (a'b)' = b'a. So b'a is in H,
and aH = bH.

***

(c) f
is onto.

***

For any x in G,
f(x'H)
= H(x')' = Hx.

***

(This is without regard to whether
G is finite or whether the number of left or right cosets is finite.)

2. (a) If s
is an element of S_{n} and s
= (a b)(b c) with a, b, c distinct, show that s
can be written as a single 3-cycle.

***

(a b)(a c) = (a b c). Just work
it out.

***

(b) If s
is an element of S_{n}and
s
= (a b)(c d) with a, b, c, d distinct, show that s
is a product of two 3-cycles by inserting e = (b c)(b c) between (a b)
and (c d).

***

(a b)(c d) = [(a b)(b c)][(b c)(c
d)] = (a c b)(b d c)

***

(c) Prove that every even permutation
in S_{n}_{ }can
be written as a product of 3-cycles. (Each individual 3-cycle is even,
so whether there is an even number of 3-cycles in the product or an odd
number is irrelevant.)

***

Every even permutation is a product
of an even number of transpositions, and each successive pair of transpositions
can be replaced by a product of 3-cycles.

***

3. Show that if H is a subgroup of
G and the index (G: H) = 2, then H is a normal subgroup of G. (Hint. As
a start you could show that for every element a not in G, aH = Ha, because
each must be the set difference G - H.)

***Because there are only two left
cosets and one of them is H, the other must be G - H.
Similarly there is one right coset other than H, and it must also
be G - H. But the other left coset is aH for some
a not in H. Then the right coset Ha is not H because
a is not in H. Therefore aH = G - H = Ha.

Finally, Let g be any element of G. Then g is either in H or in aH = Ha.

If g is in H, then so is g'. Then because H is closed under multiplication, g'Hg is a subset of H. On the other hand, if h is in H, then ghg' is in H and then h = g'(ghg')g is in g'Hg. So each of H and g'Hg is a subset of the other, and they must be the same subset of G.

Now let g be in aH
, so that g = ak for some k in H. If x
is in g'Hg, then there is an h in H
so that x = g'hg = k'a'hak = k'(a'ha)k.

But Ha = aH, so that ha
= ah_{1} for some h_{1} in H.
Then

x = k'(a'ha)k = k'(a'ah_{1})k
= k'h_{1}k, which is in H. So, g'Hg is a subset of
H. Finally if h is in H, then h = g'(ghg')g = g'(akhk'a')g.
But akhk' = a(khk') is in aH = Ha. So there is an element
h_{2} in H such that akhk' = h_{2}a.
Then h = g'(akhk'a')g = g'(h_{2}aa')g = g'h_{2}g,
which is in g'Hg. So H is a subset of g'Hg.
Hence, H = g'Hg.

So for all g in G,
H = g'Hg, and H is a normal subgroup of G.

***

4. Let G and C be groups and let F
be a homomorphism of G into C. Let a be any element of G and let k be any
integer. Let e be the identity in G and f be the identity in C. We proved
in class that F(e) = f. Since a^{0} = e and F(a)^{0} =
f, we now know that F(a^{0}) = F(a)^{0}.

(a) Prove by induction that if k is
positive then F(a^{k}) = F(a)^{k}.

***If k = 1, then F(a^{1})
= F(a) = F(a)^{1}. If F(a^{k}) = F(a)^{k
}then

F(a^{k + 1}) = F(a^{k}a)
= F(a^{k})F(a) = F(a)^{k}F(a)^{ }=^{ }
F(a)^{k + 1}.

***

(b) We proved in class that F(a¢
) = F(a)¢ . Put
this together with part (a) to deduce that if k is negative, then F(a^{k})
= F(a)^{k}.

***If k is negative, then
k = -m for some positive m. Then

F(a^{k}) = F(a^{-m})
= F((a')^{m}) = F(a')^{m }= [F(a)']^{m}.
= F(a)^{-m }= F(a)^{k}
^{***}

[You've now shown formally that for
any integer k, F(a^{k}) = F(a)^{k}.]

5. Let G = (g) be a cyclic group. Our
main unresolved question about cyclic groups is to determine which elements
g^{k} other than g itself can be generators of G. Another way to
phrase this is to observe that for every integer k (positive or negative)
we get a subgroup (g^{k}) of G.

Then the question becomes for
which values of k is (g^{k}) = G?

(a) Show that if G is infinite then
(g^{k}) = G if and only if k = 1(trivial) or k = -1.

***

If k = -1, then g
is the inverse of g^{-1 }and g is in (g^{-1}).
So, all powers of g are in (g^{-1}). So
(g^{-1}) = G.

On the other hand, if (g^{k})
= G for some integer k, then g is a power of g^{k}.
So, g^{1} = g = (g^{k})^{m }= g^{km} for
some integer m. But in an *infinite* cyclic group (g),
different powers of the generator g cannot be the same group
element. Therefore, 1 = km. But the only integers
k which divide 1 are k = +1 and k =
-1.

***

(b) Show that if G is finite and has
n elements, then (g^{k}) = G if and only if the greatest common
factor (n, k) = 1. (You may use everything that we did in Math 580 with
integers, factors, and greatest common factors.)

***

If (g^{k})
= G then g is an element of (g^{k}).
Then g^{1} = g = (g^{k})^{m }= g^{km }
for some integer m. But we know that in a finite cyclic group
of order n, g^{n} = e and g^{i}
= g^{j }if and only if i - j is a multiple of
n. Hence 1 - km = ns for some integer s.
So,

1 = ns + km, and k and
n are relatively prime.

Conversely if k and
n are relatively prime then 1 = ns + km for some
s and m. then g = g^{1} = (g^{n})^{s}(g^{k})^{m
}=
(e^{s})(g^{k})^{m} = (g^{k})^{m}.
Therefore, g is an element of (g^{k}), and
g^{k} generates G.

***