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.