Primary Reference Title: Algebra by Thomas W. Hungerford Series: Graduate Texts in Mathematics (Volume 73), Springer Scope: Selected exercise solutions and theoretical proofs.

Chapter 1: Groups

Section 1.1: Semigroups, Monoids and Groups

Exercise (1.1.1). Give examples other than those in the text of semigroups and monoids that are not groups.

Solution.

Take $S = {a, b, c}$ and define the binary operation as $x y = x$ for any $x, y \in S$ . This operation is associative, but the set does not have a two-sided identity. So, $(S, \cdot)$ is a semigroup but not a monoid.
Let $ℬ = {0, 1}$ and define the operation as follows:

$+$	$0$	$1$
$0$	$0$	$1$
$1$	$1$	$1$

$(ℬ, +)$ is a monoid (with identity $0$ ), but the element $1$ does not have an inverse, hence it is not a group.

Exercise (1.1.2). Let $G$ be a group (written additively), $S$ a nonempty set, and $M (S, G)$ the set of all functions $f : S \to G$ . Define addition in $M (S, G)$ as $(f + g) (s) = f (s) + g (s)$ . Prove that $M (S, G)$ is a group, which is abelian if $G$ is.

Solution. 1. Non-emptiness: $M (S, G)$ is not empty since the constant function $f (s) = 0_{G}$ for all $s \in S$ is in $M (S, G)$ .

2. Associativity: Take $f, g, h \in M (S, G)$ . For any $s \in S$ : $[f + (g + h)] (s) = f (s) + (g (s) + h (s)) = (f (s) + g (s)) + h (s) = [(f + g) + h] (s)$

3. Identity: The identity element is the constant function $f (s) = 0_{G}$ . For any $g \in M (S, G)$ : $(f + g) (s) = 0_{G} + g (s) = g (s) = g (s) + 0_{G} = (g + f) (s)$

4. Inverse: For every $f \in M (S, G)$ , define $g (s) = - f (s)$ . Since $G$ is a group, $- f (s)$ is well-defined in $G$ : $(f + g) (s) = f (s) + (- f (s)) = 0_{G}$

Thus, $f + g$ is the identity function. As a result, $M (S, G)$ is a group. If $G$ is abelian, then $f (s) + g (s) = g (s) + f (s)$ , making $M (S, G)$ abelian as well. $◼$

Exercise (1.1.4). If $G$ is a group such that $x^{2} = e$ for all $x \in G$ , prove that $G$ is abelian.

Proof. Let $a, b \in G$ . We want to show that $a b = b a$ . By the given hypothesis, for any element in the group, its square is the identity $e$ . Since $G$ is closed under its binary operation, the product $a b \in G$ . Therefore, we have:

$(a b)^{2} = e ⟹ a b a b = e$

To isolate the terms, we can multiply the equation on the left by $a$ and on the right by $b$ :

$a (a b a b) b = a e b ⟹ (a^{2}) b a (b^{2}) = a b$

Since $a^{2} = e$ and $b^{2} = e$ by our initial hypothesis, the equation simplifies to:

$e b a e = a b ⟹ b a = a b$

Since $a$ and $b$ were chosen arbitrarily from $G$ , the group operation is commutative. Thus, $G$ is an abelian group. $◼$