**1.**
(1.3, p 41, #5)

**Comment:**
You might try to use a counting argument to show that HK
has the same number of elements as Z^x_n.
Or you could try to work with a related homomorphism.

This can also be treated as a special case of 1.4 #26
(but that isn't particularly easy either).

**Corrected version:**

**2.**
(1.3, p 41, #8)

This exercise concerns subgroups of $\Z \times \Z$.

(a) For each positive integer $n>1$, let

$C_n = \{ (a,b) \in \Z \times \Z | a \equiv b \mod{n} \}$.
Show that $C_n$ is a subgroup of $\Z \times \Z$.

(b) Let $D = \{ (a,a) | a \in \Z \}$
be the ``diagonal'' subgroup of $\Z \times \Z$.
Show that every subgroup of $\Z \times \Z$ that contains $D$
has the form $C_n$, for some positive integer $n$.

(c) Show that $C_n \cong \Z \times \Z$.

**Comment:**
To get an idea of what C_n will be,
you might start writing out all of the elements of C_5.
They should be in 1-1 correspondence with Z x Z;
now what function will produce an isomorphism?

**3.**
(1.3, p 42, #10)

**Comment:**
On Friday I wrote out the possibilities for S_4.

There are 4-cycles of the form (a,b,c,d),
and there should be 4! / 4 = 6 of these.
Remember that (a,b,c,d)=(b,c,d,a)=(c,d,a,b)=(d,a,b,c).

There are \frac{4 \cdot 3 \cdot 2}{3} = 8 3-cycles (a,b,c).

There are \frac{4 \cdot 3}{2} = 6 2-cycles (a,b).

There are \frac{4 \cdot 3}{2} \frac{2 \cdot 1}{2} \frac{1}{2} = 3
cycles of the form (a,b)(c,d).

Finally, with the identity, we have a total of 1+3+6+8+6=24 elements.

Of these, the identity, and permutations of the form (a,b)(c,d) and (a,b,c)
are even, so |A_4| = 12.

**4.**
(1.3, p 42, #20)

**Comment:**
Let's look at S_3, in the form { 1,a,a^2,b,ab,a^2b}.

Then A_3 = {1,a,a^2 }, and these are squares since a = (a^2)^2.
These are the only squares since b^2 = (ab)^2 = (a^2b)^2 = 1.

**5.**
(1.4, p 54, #5)

**6.**
(1.4, p 54, #13)

**7.**
(1.4, p 55, #17)

**Comment:**
It isn't practical to examine all possible one-to-one correspondences
to show that none will produce an isomorphism.
You need to find some algebraic property of one group
(that you know would be preserved by an isomorphism)
that is not shared by the second.
Here you might think about orders of elements,
or orders of subgroups.

**8.**
(1.4, p 54, #25)

**Hint:**
To get any information at all out of the assumption,
you have to be able to construct various automorphisms
and then draw some conclusions because these automorphisms
have to be the identity mapping phi (x) = x for all x in G.
Look at problem #15 in Section 1.1.

**9.**
(1.4, p 54, #27)

**10.**
(1.4, p 54, #29)

**Comment:**
You can't assume that the group is finite--do some
examples in Z.