**Problem 1.1.** Let
be a set. Prove that the law of composition defined by
for all
and
in
is associative. For which sets does this law have an
identity?

*Solution.* Given
, one has
so this law of composition on
is associative. Next, I claim that this law has an
identity element if and only if
is a singleton set. Indeed, if
, then
, so
is the identity element. Conversely, suppose
contain an identity element
. Then, for any
, one has
. But the law of composition implies that
, hence
and
.

**Problem 1.3.** Let
denote the set
of natural numbers, and let
be the *shift* map defined by
. Prove that
has no right inverse, but that it has infinitely many left
inverses.

*Solution.* Suppose, for the sake of contradiction, that there
exists a right inverse
for
. Then
for every
. In particular, one has
, from which it follows that
in which case
, which is a contradiction. So no such right inverse
exists.

Now, fix , and define by Then, for any , one has so is a left inverse of . As can be chosen arbitrarily from , we have in fact defined a countable family of left inverses.

**Problem 2.1.** Make a multiplication table for the
symmetric group
.

*Solution.* The standard presentation for the symmetric group
, given in (2.2.7) in Artin, is
where
and
. Using these defining relations, we compute the
multiplication table for
.

**Problem 4.3.** Let
and
be elements of a group
. Prove that
and
have the same order.

*Solution.* We begin with a lemma.

*Lemma.* For any
,
*Proof.* We proceed by induction on
. When
, we find that
. Suppose inductively that
for some fixed
. It then follows that
which closes the induction.

Now, for any , we have if and only if , which is true if and only if , which is true if and only if . So has finite order if and only if has finite order, hence has infinite order if and only if has infinite order. Furthermore, if and have finite order, then the forward direction implies that the order of divides the order of and the backward direction that the order of divides the order of . As and are positive, we then conclude that .

**Problem 5.1.** Let
be a surjective homomorphim. Prove that if
is cyclic, then
is cyclic, and if
is abelian, then
is abelian.

*Solution.* Suppose that
is cyclic. So there exists
such that
. Fix
. As
is surjective, there exists
such that
. As
is cyclic, there exists
such that
. Then
so
, hence
. As
by definition, we conclude that
, so
is cyclic.

Next, let . By surjectivity of , there exist such that and . As is abelian, we find that so is abelian, as required.

**Problem 5.2.** Prove that the intersection of
of subgroups of a group
is a subgroup of
, and that if
is a normal subgroup of
, then
is a normal subgroup of
.

*Proof.* We have
by definition
. Let
. Then
and
. As
and
are closed under composition, we have
and
, so
. As
and
are subgroups, they contain the identity element of
, so
. Finally, given
, there exists an inverse
. As
and
are closed under inversion, we have
and
, so
. So
is a subgroup of
.

Suppose now that is a normal subgroup of . Let and . Then , , and , so . Furthermore, one has so because is a normal subgroup of , we have . So , so is a normal subgroup of .

**Problem 6.1.** Let
be the group of real matrices of the form
. Is the map
that sends
to this matrix an isomorphism?

*Solution.* Define
by
. I claim that
is an isomorphism. First, if
for
, then
. Equating entries then yields
, so
is injective. Second, given
, we have
, so
is surjective and hence bijective. Finally, we will show
that
is a homomorphism. For any
, we find that
so
is a homomorphism and hence an isomorphism, hence
, as required.

**Problem 6.4.** Prove that in a group, the products
and
are conjugate elements.

*Solution.* Let
. We have
where
, so
and
are conjugate elements.