Academic Notes

📚 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 $xy = x$ 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 $\mathcal{B} = \{0, 1\}$ and define the operation as follows:
    + 0 1
    0 0 1
    1 1 1
    $(\mathcal{B}, +)$ 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.

Proof / 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. $\blacksquare$

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 $ab = ba$. 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 $ab \in G$. Therefore, we have:

$$ (ab)^2 = e \implies abab = e $$

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

$$ a(abab)b = aeb \implies (a^2)ba(b^2) = ab $$

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

$$ ebae = ab \implies ba = ab $$

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