A group is a set $G$ together with a binary operation $\circ$. The binary operation $\circ$ takes two elements of $G$, and returns an element of $G$. We write $x \circ y$ to indicate the returned value of $\circ$ on input $x$ and $y$. Note that the ordering is important: $x \circ y$ might not be equal to $y \circ x$. If you like, you can instead think of a function $b : G \times G \rightarrow G$ where $x \circ y$ is just shorthand for $b(x,y)$. The binary operation must satisfy:
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Problem Ⅲ: Group Theory (You are currently browsing this post)
Proposition 3.a.1. Let $G = {a, b}$. Define a binary operation $\circ : G \times G \to G$ by prescribing the following values:
\[a \circ a = a, \quad a \circ b = b, \quad b \circ a = b, \quad b \circ b = a.\]Then the algebraic structure $(G, \circ)$ is a group.
Proof. Consider the set $G = {a, b}$ and the operation $\circ : G \times G \to G$ defined by
\[\forall x,y \in G: x \circ y = \begin{cases} a, & \text{if } (x,y) = (a,a) \text{ or } (b,b), \\[6pt] b, & \text{if } (x,y) = (a,b) \text{ or } (b,a). \end{cases}\]Closure: By construction of the operation $\circ$, for each ordered pair $(x,y) \in G \times G$, the element $ x \circ y $ is defined to be either $ a $ or $ b $, both of which lie in $ G $. Hence:
\[\forall x,y \in G: x \circ y \in G.\]Associativity: To establish that $\circ$ is associative, it suffices to verify that for all $ x,y,z \in G $, the equality
\[(x \circ y) \circ z = x \circ (y \circ z)\]holds. Since $\vert G \vert =2$, one must verify all possible combinations of $(x,y,z)$. The tuples in $G \times G \times G$ are $(a,a,a), (a,a,b), (a,b,a), (a,b,b), (b,a,a), (b,a,b), (b,b,a), (b,b,b)$.
Consider each case in turn:
For $(x,y,z) = (a,a,a)$:
\[(a \circ a) \circ a = a \circ a = a, \quad a \circ (a \circ a) = a \circ a = a.\]For $(x,y,z) = (a,a,b)$:
\[(a \circ a) \circ b = a \circ b = b, \quad a \circ (a \circ b) = a \circ b = b.\]For $(x,y,z) = (a,b,a)$:
\[(a \circ b) \circ a = b \circ a = b, \quad a \circ (b \circ a) = a \circ b = b.\]For $(x,y,z) = (a,b,b)$:
\[(a \circ b) \circ b = b \circ b = a, \quad a \circ (b \circ b) = a \circ a = a.\]For $(x,y,z) = (b,a,a)$:
\[(b \circ a) \circ a = b \circ a = b, \quad b \circ (a \circ a) = b \circ a = b.\]For $(x,y,z) = (b,a,b)$:
\[(b \circ a) \circ b = b \circ b = a, \quad b \circ (a \circ b) = b \circ b = a.\]For $(x,y,z) = (b,b,a)$:
\[(b \circ b) \circ a = a \circ a = a, \quad b \circ (b \circ a) = b \circ b = a.\]For $(x,y,z) = (b,b,b)$:
\[(b \circ b) \circ b = a \circ b = b, \quad b \circ (b \circ b) = b \circ a = b.\]In each of these eight cases, the left-hand side $(x \circ y) \circ z$ equals the right-hand side $x \circ (y \circ z)$. Therefore, associativity holds:
\[\forall x,y,z \in G: (x \circ y) \circ z = x \circ (y \circ z).\]Existence of an Identity Element: An identity element $ i \in G $ must satisfy
\[\forall x \in G: x \circ i = x \quad \text{and} \quad i \circ x = x.\]By inspection of the defining relations of $\circ$, the element $ a $ behaves as an identity:
\[a \circ a = a, \quad a \circ b = b, \quad b \circ a = b.\]Hence:
\[\forall x \in G: x \circ a = x \quad \text{and} \quad a \circ x = x.\]Thus, $ a $ serves as the identity element $ i $ of $(G, \circ)$.
Existence of Inverses: For each $ x \in G $, an inverse $ x^{-1} \in G $ must satisfy
\[x \circ x^{-1} = i \quad \text{and} \quad x^{-1} \circ x = i.\]Since the identity element $ i = a $ has been identified, one must determine inverses with respect to $ a $:
For $ x = a $:
\[a \circ a = a = i, \quad a \circ a = i.\]Thus, $ a^{-1} = a $.
For $ x = b $:
\[b \circ b = a = i, \quad b \circ b = i.\]Thus, $ b^{-1} = b $.
Therefore:
\[a^{-1} = a, \quad b^{-1} = b.\]Since all four group axioms—closure, associativity, existence of an identity, and existence of inverses—have been rigorously verified, it follows that $(G, \circ)$ is indeed a group. $\textbf{Q.E.D.}$
Commentary
The constructed group $(G,\circ)$ is isomorphic to the cyclic group of order 2, often denoted by $\mathbb{Z}/2\mathbb{Z}$, under addition modulo 2.
Proposition 3.b.1. Let $G = { i, a, b, c }$ be a set of four distinct elements, and define a binary operation $\circ : G \times G \to G$ by prescribing the following properties:
\[i \circ i = i, \quad a \circ a = i, \quad b \circ b = i, \quad c \circ c = i,\]and
\[a \circ b = c, \quad b \circ a = c, \quad b \circ c = a, \quad c \circ b = a, \quad c \circ a = b, \quad a \circ c = b.\]Furthermore, for every $x \in G$,
\[i \circ x = x \quad \text{and} \quad x \circ i = x.\]Then $(G, \circ)$ is a group of order 4, and each of its elements satisfies the property $x \circ x = i$.
Proof. Define $G = {i, a, b, c}$ and the operation $\circ : G \times G \rightarrow G$ by the Cayley table:
\[\begin{array}{c|cccc} \circ & i & a & b & c \\ \hline i & i & a & b & c \\ a & a & i & c & b \\ b & b & c & i & a \\ c & c & b & a & i \end{array}\]Closure: By inspection of the table, for all $x, y \in G$, the element $x \circ y$ is always one of $i, a, b, c$. Thus:
\[\forall x,y \in G:\; x \circ y \in G.\]Associativity: To demonstrate associativity, it suffices to verify that for all $x,y,z \in G$,
\[(x \circ y) \circ z = x \circ (y \circ z).\]Since $\vert G \vert =4$, a direct verification can be conducted by enumerating all triplets $(x,y,z)$. Each computation conforms to the pattern of a direct product of two-element groups or can be seen to follow from the algebraic structure of the Klein four-group, commonly denoted by $V_4$ or $C_2 \times C_2$, which is known to be associative.
Notably, the structure defined by these rules is isomorphic to the Klein four-group, which is well known in group theory to be associative. Hence, no contradictions arise, ensuring:
\[\forall x,y,z \in G:\; (x \circ y) \circ z = x \circ (y \circ z).\]Identity Element: By examination of the table, the element $i$ serves as an identity, since:
\[\forall x \in G:\; x \circ i = x \quad \text{and} \quad i \circ x = x.\]Thus $i$ is the identity element of the group $(G,\circ)$.
Inverses: Each element is its own inverse. By construction:
\[a \circ a = i, \quad b \circ b = i, \quad c \circ c = i, \quad i \circ i = i.\]Therefore, $a^{-1} = a$, $b^{-1} = b$, $c^{-1} = c$, and $i^{-1} = i$.
Since all group axioms—closure, associativity, identity, and inverses—are satisfied, $(G,\circ)$ is a group. $\textbf{Q.E.D.}$
Commentary
The structure $(G,\circ)$ thus obtained is known as the Klein four-group, and it is a classic example of a group in which every element is of order 2, except the identity.
Proposition 3.c.1. Let $G = {0,1,2,3}$ and define a binary operation $\circ : G \times G \to G$ by
\[x \circ y = (x + y) \mod 4,\]for all $x,y \in G$, where $+$ denotes the usual addition of integers and $\mod 4$ denotes the remainder upon division by 4. Then $(G, \circ)$ is a group, and there exists an element $x \in G$ such that $x \circ x \neq i$, where $i$ is the identity element.
Proof. Consider the set $G = {0,1,2,3}$. Define the binary operation $\circ : G \times G \to G$ by the rule
\[x \circ y = (x + y) \bmod 4.\]Closure: Since $x,y \in {0,1,2,3}$, the integer sum $x + y$ is an integer between 0 and 6. Taking this sum modulo 4 yields an element in ${0,1,2,3}$. Thus:
\[\forall x,y \in G, \quad x \circ y \in G.\]Associativity: For any $x,y,z \in G$, we must verify
\[(x \circ y) \circ z = x \circ (y \circ z).\]By definition:
\[(x \circ y) \circ z = ((x + y) \bmod 4 + z) \bmod 4,\] \[x \circ (y \circ z) = (x + (y + z) \bmod 4) \bmod 4.\]Since addition modulo 4 is associative, it follows that:
\[((x + y) + z) \bmod 4 = (x + (y + z)) \bmod 4.\]Therefore, associativity holds:
\[\forall x,y,z \in G, \; (x \circ y) \circ z = x \circ (y \circ z).\]Identity Element: An identity element $i \in G$ must satisfy
\[\forall x \in G, \; x \circ i = x \quad \text{and} \quad i \circ x = x.\]By inspection, when choosing $i = 0$:
\[x \circ 0 = (x + 0) \bmod 4 = x,\] \[0 \circ x = (0 + x) \bmod 4 = x.\]Thus, $i = 0$ is the identity element of $(G,\circ)$.
Inverses: For each $x \in G$, we must find $x^{-1} \in G$ such that
\[x \circ x^{-1} = 0, \quad x^{-1} \circ x = 0,\]where $0$ is the identity. Since addition modulo 4 forms a cyclic group, each element is invertible:
For $x = 0$:
\[0 + 0 \equiv 0 \pmod{4}, \text{ so } 0^{-1} = 0.\]For $x = 1$:
\[1 + 3 \equiv 0 \pmod{4}, \text{ so } 1^{-1} = 3.\]For $x = 2$:
\[2 + 2 \equiv 0 \pmod{4}, \text{ so } 2^{-1} = 2.\]For $x = 3$:
\[3 + 1 \equiv 0 \pmod{4}, \text{ so } 3^{-1} = 1.\]Hence, every element has an inverse.
Since $(G, \circ)$ satisfies closure, associativity, has an identity element, and each element has an inverse, it follows that $(G, \circ)$ is a group.
Consider $x = 1$. Then:
\[1 \circ 1 = (1 + 1) \bmod 4 = 2.\]Note that $i = 0$, thus:
\[1 \circ 1 = 2 \neq 0.\]Thus, there exists an element in $G$, namely $x = 1$, for which $x \circ x \neq i$. In fact, in this group, only the element $2$ satisfies $x \circ x = i$, whereas $1$ and $3$ do not. $\textbf{Q.E.D.}$
Commentary
This confirms the existence of a group of order 4 in which not every element is an involution.
Proposition 3.d.1. Let $(G,\circ)$ be a group with identity element $i$. Suppose $x,y,z \in G$ and assume that
\[x \circ y = x \circ z.\]Then $y = z$.
Proof. Since $(G,\circ)$ is a group, each element $x \in G$ has an inverse $x^{-1} \in G$ such that
\[x^{-1} \circ x = x \circ x^{-1} = i,\]where $i$ is the identity element of $G$.
Given the equation
\[x \circ y = x \circ z,\]apply the inverse $x^{-1}$ to the left of both sides:
\[x^{-1} \circ (x \circ y) = x^{-1} \circ (x \circ z).\]By associativity of the group operation $\circ$, we have:
\[(x^{-1} \circ x) \circ y = (x^{-1} \circ x) \circ z.\]Since $x^{-1} \circ x = i$, it follows that:
\[i \circ y = i \circ z.\]By the defining property of the identity element $i$:
\[y = z.\]This completes the proof, showing that left-cancellation holds in a group. $\textbf{Q.E.D.}$
Proposition 3.e.1. Let $(G,\circ)$ be a group with identity element $i$, and let $x \in G$ be arbitrary. Suppose there is an element $x’ \in G$ such that
\[x' \circ x = i.\]Then $x’ = x^{-1}$, where $x^{-1}$ is the unique inverse of $x$.
Proof. By the definition of the inverse, the element $x^{-1} \in G$ is characterized by:
\[x \circ x^{-1} = i \quad \text{and} \quad x^{-1} \circ x = i.\]To establish uniqueness, assume that $x’$ is any element satisfying
\[x' \circ x = i.\]Consider the element $x^{-1}$ and the given equality $x’ \circ x = i$. Pre-multiplying both sides by $x^{-1}$, we obtain:
\[x^{-1} \circ (x' \circ x) = x^{-1} \circ i.\]By associativity:
\[(x^{-1} \circ x') \circ x = x^{-1} \circ i.\]Since $x^{-1} \circ i = x^{-1}$, it follows that:
\[(x^{-1} \circ x') \circ x = x^{-1}.\]Now consider the known identity $x \circ x^{-1} = i$. We rewrite the last equation in a form that will allow us to use this identity. Since $(x^{-1} \circ x’) \circ x = x^{-1}$, right-multiplying both sides by $x^{-1}$ gives:
\[(x^{-1} \circ x') \circ (x \circ x^{-1}) = x^{-1} \circ x^{-1}.\]As $x \circ x^{-1} = i$, we have:
\[(x^{-1} \circ x') \circ i = x^{-1} \circ x^{-1}.\]Since $y \circ i = y$ for all $y \in G$, it follows:
\[x^{-1} \circ x' = x^{-1} \circ x^{-1}.\]Left-cancellation (already established from the group axioms, see Proposition 3.d.1) implies:
\[x' = x^{-1}.\]Thus, any element $x’$ that satisfies $x’ \circ x = i$ must be equal to $x^{-1}$. The inverse of any element $x \in G$ is therefore unique. $\textbf{Q.E.D.}$
Commentary
The result established in Question (d) is often referred to as the left-cancellation law, and it constitutes one of the canonical consequences of group axioms. The argument exploits the existence of inverses: given an equation of the form $x \circ y = x \circ z$, one can multiply on the left by the inverse of ( x ) to reduce the equation to $y = z$. It ensures that the group operation, though potentially non-commutative, still behaves in a manner that prevents distinct elements from collapsing into one another under the same left factor. In other words, the group structure is sufficiently rigid to forbid pathological identifications that could arise in weaker algebraic structures lacking inverses.
The uniqueness of inverses, addressed in Quesiton (e), is another pivotal consequence of the group axioms. Although the existence of an inverse for each element is assumed in the definition of a group, it is not immediately obvious that there cannot be two distinct inverses of the same element. The argument to show uniqueness again leverages the group axioms, particularly the presence of a suitable identity element and the power of cancellation. By assuming that some element $x’$ behaves as an inverse of $x$ and then employing the cancellation laws, one can verify that this $x’$ must coincide with the known inverse $x^{-1}$.
Proposition 3.f.1. Let $(G,\circ)$ be a group with identity element $i$. For any elements $x,y \in G$, the inverse of the product $x \circ y$ is given by
\[(x \circ y)^{-1} = y^{-1} \circ x^{-1}.\]Proof. Consider arbitrary elements $x,y \in G$. To establish that $(x \circ y)^{-1} = y^{-1} \circ x^{-1}$, it suffices to verify that
\[(x \circ y) \circ (y^{-1} \circ x^{-1}) = i \quad \text{and} \quad (y^{-1} \circ x^{-1}) \circ (x \circ y) = i,\]where $i$ is the identity element of the group $(G,\circ)$.
First, consider the product $(x \circ y) \circ (y^{-1} \circ x^{-1})$. By associativity of the group operation $\circ$, we have
\[(x \circ y) \circ (y^{-1} \circ x^{-1}) = x \circ (y \circ (y^{-1} \circ x^{-1})).\]Regrouping and using associativity again, we obtain
\[x \circ (y \circ (y^{-1} \circ x^{-1})) = x \circ ((y \circ y^{-1}) \circ x^{-1}).\]Since $y^{-1}$ is the inverse of $y$, we know $y \circ y^{-1} = i$. Substitute this into the equation:
\[x \circ ((y \circ y^{-1}) \circ x^{-1}) = x \circ (i \circ x^{-1}).\]Since $i$ is the identity element, $i \circ x^{-1} = x^{-1}$. Thus:
\[x \circ (i \circ x^{-1}) = x \circ x^{-1}.\]Since $x^{-1}$ is the inverse of $x$, $x \circ x^{-1} = i$. Therefore:
\[(x \circ y) \circ (y^{-1} \circ x^{-1}) = i.\]A symmetrical argument applies when we consider the product $(y^{-1} \circ x^{-1}) \circ (x \circ y)$:
\[(y^{-1} \circ x^{-1}) \circ (x \circ y) = y^{-1} \circ (x^{-1} \circ (x \circ y))\]by associativity. Reassociating:
\[y^{-1} \circ (x^{-1} \circ (x \circ y)) = y^{-1} \circ ((x^{-1} \circ x) \circ y).\]Since $x^{-1}$ is the inverse of $x$, $x^{-1} \circ x = i$. Substitute this:
\[y^{-1} \circ ((x^{-1} \circ x) \circ y) = y^{-1} \circ (i \circ y).\]The identity property $i \circ y = y$ gives:
\[y^{-1} \circ (i \circ y) = y^{-1} \circ y = i.\]Thus,
\[(y^{-1} \circ x^{-1}) \circ (x \circ y) = i.\]We have shown both
\[(x \circ y) \circ (y^{-1} \circ x^{-1}) = i \quad \text{and} \quad (y^{-1} \circ x^{-1}) \circ (x \circ y) = i.\]By definition of the inverse, this demonstrates that $y^{-1} \circ x^{-1}$ is indeed the inverse of $x \circ y$. Since inverses are unique, it follows that:
\[(x \circ y)^{-1} = y^{-1} \circ x^{-1}.\]$\textbf{Q.E.D.}$
Commentary
This property is mirrored in other algebraic structures (like rings or modules) and has analogues in categorical settings.
Proposition. 3.g.1 Let $(G, \circ)$ be a group with identity element $i$. Then $G$ is abelian if and only if for all $x, y \in G$,
\[x \circ y \circ x^{-1} \circ y^{-1} = i.\]Proof. Necessity: Suppose $G$ is abelian. By definition, for all $x, y \in G$,
\[x \circ y = y \circ x.\]We aim to show that under this condition,
\[x \circ y \circ x^{-1} \circ y^{-1} = i.\]Consider the left-hand side:
\[x \circ y \circ x^{-1} \circ y^{-1}.\]Since $G$ is abelian, $x \circ y = y \circ x$. Thus,
\[x \circ y \circ x^{-1} \circ y^{-1} = y \circ x \circ x^{-1} \circ y^{-1}.\]Using the property that $x \circ x^{-1} = i$,
\[y \circ x \circ x^{-1} \circ y^{-1} = y \circ i \circ y^{-1}.\]Since $i$ is the identity element,
\[y \circ i \circ y^{-1} = y \circ y^{-1}.\]Again, by the definition of inverses,
\[y \circ y^{-1} = i.\]Therefore,
\[x \circ y \circ x^{-1} \circ y^{-1} = i.\]This concludes the necessity part of the proof.
Sufficiency: Conversely, assume that for all $x, y \in G$,
\[x \circ y \circ x^{-1} \circ y^{-1} = i.\]We aim to show that $G$ is abelian, i.e., $x \circ y = y \circ x$ for all $x, y \in G$.
Start by manipulating the given equation:
\[x \circ y \circ x^{-1} \circ y^{-1} = i.\]We can rearrange the terms using associativity:
\[(x \circ y) \circ (x^{-1} \circ y^{-1}) = i.\]Recall that in a group, the inverse of a product satisfies:
\[(x \circ y)^{-1} = y^{-1} \circ x^{-1}.\]Thus, substituting,
\[(x \circ y) \circ (x \circ y)^{-1} = i.\]This is an identity in any group, confirming that $(x \circ y)^{-1}$ is indeed the inverse of $x \circ y$. However, this does not directly aid in establishing commutativity.
To proceed, consider pre-multiplying both sides of the original assumption by $y \circ x$:
\[(y \circ x) \circ (x \circ y \circ x^{-1} \circ y^{-1}) = y \circ x \circ i.\]Simplifying the right-hand side,
\[y \circ x \circ i = y \circ x.\]On the left-hand side, by associativity,
\[(y \circ x) \circ (x \circ y \circ x^{-1} \circ y^{-1}) = (y \circ x \circ x \circ y) \circ x^{-1} \circ y^{-1}.\]Notice that $x \circ x^{-1} = i$ and $y \circ y^{-1} = i$, but the manipulation becomes non-trivial. Instead, a more straightforward approach involves using the given condition to directly derive commutativity.
Consider the equation:
\[x \circ y \circ x^{-1} \circ y^{-1} = i.\]Multiply both sides on the right by $y$:
\[x \circ y \circ x^{-1} \circ y^{-1} \circ y = i \circ y.\]Simplifying,
\[x \circ y \circ x^{-1} \circ (y^{-1} \circ y) = y.\]Since $y^{-1} \circ y = i$,
\[x \circ y \circ x^{-1} \circ i = y.\]And $x^{-1} \circ i = x^{-1}$,
\[x \circ y \circ x^{-1} = y.\]Now, pre-multiply both sides by $x^{-1}$:
\[x^{-1} \circ (x \circ y \circ x^{-1}) = x^{-1} \circ y.\]Using associativity,
\[(x^{-1} \circ x) \circ y \circ x^{-1} = x^{-1} \circ y.\]Since $x^{-1} \circ x = i$,
\[i \circ y \circ x^{-1} = x^{-1} \circ y.\]And $i \circ y = y$,
\[y \circ x^{-1} = x^{-1} \circ y.\]Finally, multiply both sides on the right by $x$:
\[y \circ x^{-1} \circ x = x^{-1} \circ y \circ x.\]Simplifying,
\[y \circ i = x^{-1} \circ y \circ x.\]Thus,
\[y = x^{-1} \circ y \circ x.\]Pre-multiplying both sides by $x$,
\[x \circ y = (x \circ x^{-1}) \circ y \circ x.\]Since $x \circ x^{-1} = i$,
\[x \circ y = i \circ y \circ x = y \circ x.\]Therefore, $x \circ y = y \circ x$ for all $x, y \in G$, establishing that $G$ is abelian. $\textbf{Q.E.D.}$
Commentary
Commutators can be utilized to construct the derived subgroup of a group, which is the smallest normal subgroup generated by all commutators, denoted by $G’$. The existence of the derived subgroup $G’$ ensures the feasibility of constructing normal subgroups and quotient groups within group theory.
For any group $G$, if its derived subgroup $G’$ is the trivial group, then $G$ must be an abelian group. This is because the generators of $G’$ are all commutators, and when \(G' = \{i\}\), all commutators equal the identity element. Consequently, $G$ satisfies the commutativity condition.