Groups
Definition 1. A binary operation on a set \(S\) is a function \(*: S\times S \to S\). For \(a,b\in S\), we denote \(*(a,b)\) by \(a*b\), or simply \(ab\). If \(*\) is a binary opeation on \(S\) and \(A\subseteq S\), then \(A\) is closed under \(*\) if for each \(a,b\in A\) one has \(a*b\in A\).
Definition 2. Let \(*\) be a binary operation on a set \(S\).
We call \(*\) associative if \[\forall a,b,c\in S \quad (a*b)*c = a*(b*c)\]
We call \(*\) commutative if \[\forall a,b\in S \quad a*b = b*a\]
Definition 3. A group is a set \(G\) with a binary operation \(* : G\times G \to G\) satisfying the following.
The operation \(*\) is associative.
There is \(e\in G\) such that for any \(g\in G\) one has \(g*e=e*g=g\). Such an \(e\) is an identity of \(G\).
For each \(g\in G\) there is \(g^{-1}\in G\) such that \(g*g^{-1}=g^{-1}*g=e\) where \(e\) is an identity of \(G\). Such a \(g^{-1}\) is an inverse of \(g\).
Theorem 1. If \(G\) is a group, the identity \(e\in G\) is unique.
Proof. Let \(G\) be a group. Suppose \(e_1\in G\) and \(e_2\in G\) are identities. Then \(e_1 = e_1 * e_2 = e_2\). ◻
Theorem 2. If \(G\) is a group, for each \(g\in G\) the inverse \(g^{-1}\in G\) is unique.
Proof. Let \(G\) be a group. Let \(g\in G\). Suppose \(g_1\in G\) and \(g_2\in G\) are inverses of \(g\). Then \[g_1 = e * g_1 = (g_2 * g) * g_1 = g_2 * (g * g_1) = g_2 * e = g_2.\] ◻
Definition 4. If \(G\) is a group and \(H\subseteq G\) is a group with the same binary operation as \(G\), then \(H\) is a subgroup of \(G\), denoted \(H\leq G\).
Theorem 3. If \(G\) is a group and \(H\subseteq G\), then \(H\leq G\) if and only if \(e\in H\) and for all \(x,y\in H\) one has \(xy^{-1}\in H\).
Proof. Let \(G\) be a group.
Suppose \(H\leq G\). Then \(H\subseteq G\) is a group with the same binary operation as \(G\). Let \(x,y\in H\). Since \(H\) is a group, \(y^{-1}\in H\). Since \(H\subseteq G\), we have \(y,y^{-1}\in G\). Then \(y y^{-1} = e\). Since \(H\) is a group, we have \(e\in H\) and \(xy^{-1}\in H\).
Suppose \(H\subseteq G\), \(e\in H\), and for all \(x,y\in H\) one has \(xy^{-1}\in H\). Let \(a\in H\). Since \(H\subseteq G\), we have \(a\in G\). Then \(ae=ea=a\) and \(e\) is an identity of \(H\). Since \(e,a\in H\), we have \(a^{-1} = ea^{-1} \in H\). Then \(H\) has inverses. Since the operation on \(G\) is associative and \(H\subseteq G\), it follows that the operation on \(G\) is associative on \(H\). Then \(H\leq G\). ◻
Example 1. For any group \(G\), we have \(G\leq G\) and \(\{e\}\leq G\). The latter is called the trivial subgroup.
Theorem 4. Let \(G\) be a group and let \(\mathcal{H}\) be a nonempty collection of subgroups of \(G\). Then \(\cap\mathcal{H}\leq G\).
Proof. Since each \(H\in\mathcal{H}\) is a subgroup of \(G\), we have \(e\in H\). Then \(e\in\cap\mathcal{H}\). Let \(x,y\in\cap\mathcal{H}\). Then \(x,y\in H\) for each \(H\in\mathcal{H}\). Since each \(H\in\mathcal{H}\) is a subgroup of \(G\), we have \(xy^{-1}\in H\). Then \(xy^{-1}\in\cap\mathcal{H}\). It follows that \(\cap\mathcal{H}\leq G\). ◻
Definition 5. Let \(G\) be a group. Let \(S\subseteq G\). The subgroup generated by \(S\), denoted \(\langle S\rangle\), is the intersection of all subgroups of \(G\) containing \(S\). That is \[\langle S \rangle = \bigcap \{H\leq G : S\subseteq H\}\] If \(S\) is a finite set, we may denote \(\langle S\rangle\) by listing the elements of \(S\) surrounded by angle brackets and separated by commas.
Example 2. If \(S = \{x\}\), we write \(\langle x\rangle\) in place of \(\langle S\rangle\). If \(S=\{x,y\}\), we write \(\langle x, y\rangle\) in place of \(\langle S\rangle\).
Theorem 5. Let \(G\) be a group. Let \(S\subseteq G\) by nonempty. Let \(S^{-1} = \{s^{-1} : s\in S\}\). Then \(\langle S\rangle\) is the set of all products of elements in \(S\cup S^{-1}\) using the operation in \(G\). \[\langle S\rangle = \bigcup_{n=1}^\infty \left\{\prod_{j=1}^n a_j : a_1,\dots,a_n \in S\cup S^{-1}\right\}\]
Proof. Let \(H\) be the set of all products of elements in \(S\cup S^{-1}\) using the operation in \(G\).
Let \(s\in S\). Then \(e=ss^{-1}\in H\). Let \(x,y\in H\). Then there are positive integers \(m\) and \(n\) and elements \((x_j)_{j=1}^m\) and \((y_j)_{j=1}^n\) of \(S\cup S^{-1}\) such that \(x=\prod_{j=1}^m x_j\) and \(y=\prod_{j=1}^n y_j\). Then \(xy^{-1} = \prod_{j=1}^m x_j \prod_{j=1}^n y_{n-j+1}^{-1} \in H\).
Then \(H\leq G\). Moreover, \(S\subseteq H\) since \(H\) contains products of just one element of \(S\). Then \(\langle S\rangle \subseteq H\).
Let \(K\leq G\) be a subgroup with \(S\subseteq K\). Since \(K\) is a group, \(S^{-1}\subseteq K\), and any product of elements in \(S\cup S^{-1}\) using the operation in \(K\) is an element of \(K\). Since \(K\leq G\), the operation in \(K\) is the same as the operation in \(G\). Then \(H\subseteq K\). Since \(K\) is arbitrary, we have \(H\subseteq \langle S\rangle\). Then \(H=\langle S\rangle\). ◻
Definition 6. The order of a group \(G\) is the cardinality of the set \(G\), denoted \(|G|\). The order of an element \(x\in G\) is the smallest \(n\in\mathbb{Z}_+\) such that \(x^n=e\), denoted \(|x|\). If there is no such \(n\), we say that \(x\) has infinite order.
Theorem 6. Let \(G\) be a group. If \(x\in G\), then \(|\langle x\rangle| = |x|\).
Proof. WIP ◻
Definition 7. Let \(G\) be a group and \(H\leq G\). For each \(g\in G\), we define the left coset of \(H\) with \(g\) as the set \[gH = \{gh\in G: h\in H\}\] and the right coset of \(H\) with \(g\) as the set \[Hg = \{hg\in G: h\in H\}\] For each \(g\in G\), we also define the set \[gHg^{-1} = \{ghg^{-1} : h\in H\}\]
Lemma 7. Let \(G\) be a group and \(H\leq G\). Let \(x,y\in G\). The following are equivalent.
\(xH=yH\)
\(x\in yH\)
\(y\in xH\)
\(x^{-1}y\in H\)
Proof. We show \(1.\implies 2.\implies 3.\implies 4.\implies 1\).
(\(1.\implies 2\).) Suppose \(xH=yH\). Since \(e\in H\), we have \(x=xe\in xH=yH\).
(\(2.\implies 3\).) Suppose \(x\in yH\). Then there is \(h\in H\) with \(x=yh\). Since \(H\leq G\), we have \(h^{-1}\in H\). Then \(y=xh^{-1} \in xH\).
(\(3.\implies 4\).) Suppose \(y\in xH\). Then there is \(h\in H\) with \(y=xh\). Then \(x^{-1}y = h\in H\).
(\(4.\implies 1\).) Suppose \(x^{-1}y\in H\). Let \(h=x^{-1}y\). Let \(a\in xH\). Then there is \(h_1\in H\) such that \(a=xh_1\). Then \[a = xh_1 = y h^{-1} h_1 \in yH.\] Hence \(xH\subseteq yH\). Let \(b\in yH\). Then there is \(h_2\in H\) such that \(b=yh_2\). Then \[b = yh_2 = xh h_2 \in xH.\] Hence \(yH\subseteq xH\). It follows that \(xH=yH\). ◻
Lemma 8. Let \(G\) be a group and \(H\leq G\). Let \(x,y\in G\). The following are equivalent.
\(Hx=Hy\)
\(x\in Hy\)
\(y\in Hx\)
\(xy^{-1}\in H\)
Proof. We show \(1.\implies 2.\implies 3.\implies 4.\implies 1\).
(\(1.\implies 2\).) Suppose \(Hx=Hy\). Since \(e\in H\), we have \(x=ex\in Hx=Hy\).
(\(2.\implies 3\).) Suppose \(x\in Hy\). Then there is \(h\in H\) with \(x=hy\). Since \(H\leq G\), we have \(h^{-1}\in H\). Then \(y=h^{-1}x \in Hx\).
(\(3.\implies 4\).) Suppose \(y\in Hx\). Then there is \(h\in H\) with \(y=hx\). Since \(H\leq G\), we have \(h^{-1}\in H\). Then \(xy^{-1} = h^{-1}\in H\).
(\(4.\implies 1\).) Suppose \(xy^{-1}\in H\). Let \(h=xy^{-1}\). Let \(a\in Hx\). Then there is \(h_1\in H\) such that \(a=h_1 x\). Then \[a = h_1 x = h_1 h y \in Hy.\] Hence \(Hx\subseteq Hy\). Let \(b\in Hy\). Then there is \(h_2\in H\) such that \(b=h_2 y\). Then \[b = h_2 y = h_2 h^{-1} x \in Hx.\] Hence \(Hy\subseteq Hx\). It follows that \(Hx=Hy\). ◻
Lemma 9. Let \(G\) be a group and \(H\leq G\). Then the set of left cosets of \(H\) with elements of \(G\) has the same cardinality as the set of right cosets of \(H\) with elements of \(G\). That is \[|\{gH : g\in G\}| = |\{Hg : g\in G\}|\]
Proof. Define a function \(f:\{gH : g\in G\}\to \{Hg : g\in G\}\) by \(f(gH) = Hg^{-1}\). We first check that this function is well-defined. Suppose for \(g_1,g_2\in G\) that \(g_1 H = g_2 H\). Then \(g_1^{-1} g_2 \in H\). Then \(Hg_1^{-1} = Hg_2^{-1}\) and the function \(f\) is well-defined.
Let \(Hg\) be a right coset. We have \((g^{-1})^{-1}g^{-1}=e\in H\). Then \(Hg = H(g^{-1})^{-1}=f(g^{-1}H)\). Hence \(f\) is surjective.
Suppose for left cosets \(g_1 H\) and \(g_2 H\) that \(f(g_1 H) = f(g_2 H)\). Then \(Hg_1^{-1} = Hg_2^{-1}\). Then \(g_1^{-1}g_2 = g_1^{-1}(g_2^{-1})^{-1}\in H\). Then \(g_1 H = g_2 H\). Hence \(f\) is injective.
Then \(f\) is bijective and the proof is complete. ◻
Definition 8. Let \(G\) be a group and \(H\leq G\). The index of \(H\) in \(G\), denoted \([G:H]\) is the cardinality of the set of cosets of \(H\) with elements of \(G\). That is \[[G:H] = |\{gH : g\in G\}| = |\{Hg : g\in G\}|\]
Lemma 10. Let \(G\) be a group and \(H\leq G\). Then all cosets of \(H\) by elements of \(G\), whether they are left or right cosets, have the same cardinality as \(H\).
Proof. Part 1: We first show that all left cosets have the same cardinality.
Let \(aH\) and \(bH\) be left cosets. Define \(f:aH\to bH\) by \(f(ah)=bh\). We first check that this function is well-defined. Suppose for \(h_1,h_2\in H\) that \(ah_1 = ah_2\). Then \(h_1 = a^{-1}ah_1 = a^{-1}ah_2 = h_2\). Hence \(bh_1 = bh_2\) and \(f\) is well-defined.
Let \(y\in bH\). Then there is \(h\in H\) with \(y=bh\). Then \(y=bh=f(ah)\) and \(f\) is surjective.
Suppose for \(x,y\in aH\) that \(f(x)=f(y)\). There are \(h_1,h_2\in H\) such that \(x=ah_1\) and \(y=ah_2\). Then \(bh_1=bh_2\) and so \(h_1=b^{-1}bh_1 = b^{-1}bh_2 = h_2\). Then \(x=ah_1=ah_2=y\) and \(f\) is injective.
Then \(f\) is bijective. Hence all left cosets have the same cardinality.
Part 2: We now show that all right cosets have the same cardinality.
Let \(Ha\) and \(Hb\) be right cosets. Define \(f:Ha\to Hb\) by \(f(ha)=hb\). We first check that this function is well-defined. Suppose for \(h_1,h_2\in H\) that \(h_1 a = h_2 b\). Then \(h_1 = h_1 aa^{-1} = h_2 aa^{-1} = h_2\). Hence \(h_1 b = h_2 b\) and \(f\) is well-defined.
Let \(y\in Hb\). Then there is \(h\in H\) with \(y=hb\). Then \(y=hb=f(ha)\) and \(f\) is surjective.
Suppose for \(x,y\in Ha\) that \(f(x)=f(y)\). There are \(h_1,h_2\in H\) such that \(x=h_1 a\) and \(y=h_2 a\). Then \(h_1 b = h_2 b\) and so \(h_1 = h_1 bb^{-1} = h_2 bb^{-1} = h_2\). Then \(x = h_1 a = h_2 a = y\) and \(f\) is injective.
Then \(f\) is bijective. Hence all right cosets have the same cardinality.
Part 3: We finish by showing that left and right cosets have the same cardinality.
Let \(aH\) and \(Hb\) be left and right cosets, respectively. Define \(f:aH\to Hb\) by \(f(ah)=hb\). We first check that this function is well-defined. Suppose for \(h_1,h_2\in H\) that \(ah_1 = ah_2\). Then \(h_1 = h_2\) and so \(h_1 b = h_2 b\) and \(f\) is well-defined.
Let \(y\in Hb\). Then there is \(h\in H\) with \(y=hb\). Then \(y=hb=f(ah)\) and \(f\) is surjective.
Suppose for \(x,y\in aH\) that \(f(x)=f(y)\). There are \(h_1,h_2\in H\) such that \(x=ah_1\) and \(y=ah_2\). Then \(h_1 b = h_2 b\) and so \(h_1 = h_2\). Then \(x = a h_1 = a h_2 = y\) and \(f\) is injective.
Then \(f\) is bijective. Hence all cosets have the same cardinality. Since \(H=eH=He\) is a coset, all cosets have cardinality \(|H|\). ◻
Lemma 11. Let \(G\) be a group and \(H\leq G\). Then the relation \(\sim\) on \(G\) defined by \(x\sim y\) if and only if \(x^{-1}y\in H\) is an equivalence relation. The set of equivalence classes is the set of left cosets of \(H\) with elements of \(G\) \[\{[g] : g\in G\} = \{ gH : g\in G\}\]
Proof. Let \(x,y,z\in G\). Since \(H\leq G\), we have \(x^{-1}x=e\in H\). Then \(x\sim x\) and \(\sim\) is reflexive.
Suppose \(x\sim y\). Then \(x^{-1}y\in H\). Since \(H\leq G\), we have \(y^{-1}x = (x^{-1}y)^{-1}\in H\). Then \(y\sim x\) and \(\sim\) is symmetric.
Suppose \(x\sim y\) and \(y\sim z\). Then \(x^{-1}y\in H\) and \(y^{-1}z\in H\). Since \(H\leq G\), we have \(x^{-1}z = x^{-1}yy^{-1}z\in H\). Then \(x\sim z\) and \(\sim\) is transitive.
Then \(\sim\) is an equivalence relation on \(G\). Let \(g\in G\) and consider the equivalence class \([g]\). We have \[x \in [g] \iff x \sim g \iff x^{-1}g\in H \iff xH = gH \iff x\in gH.\] It follows that the set of equivalence classes is the set of left cosets of \(H\) with elements of \(G\). ◻
Theorem 12 (Lagrange). If \(G\) is a finite group and \(H\leq G\), then \[|G| = [G:H] \cdot |H|\] In particular, the order of \(H\) divides the order of \(G\).
Proof. Let \(G\) be a finite group with \(H\leq G\). Then \(H\subseteq G\) and \(H\) is finite. The relation \(\sim\) on \(G\) defined by \(x\sim y\) if and only if \(x^{-1}y\in H\) is an equivalence relation. The equivalence classes are left cosets of \(H\) with elements of \(G\), which all have cardinality \(|H|\). The equivalence classes form a partition of \(G\) and there are \([G:H]\) left cosets of \(H\) with elements of \(G\). It follows that \[|G| = [G:H] \cdot |H|.\] ◻
Corollary 13. If \(G\) is a finite group and \(x\in G\), then the order of \(x\) divides the order of \(G\).
Proof. Let \(G\) be a finite group and let \(x\in G\). Then \(H = \langle x\rangle\) is a subgroup of \(G\) with \(|H|=|x|\). By Lagrange’s theorem, the order of \(H\) divides the order of \(G\). Hence the order of \(x\) divides the order of \(G\). ◻
Definition 9. Let \(G\) be a group. A subgroup \(N\leq G\) is a normal subgroup if for all \(g\in G\) one has \(gN=Ng\). We denote this as \(N\trianglelefteq G\).
Theorem 14. Let \(G\) be a group and \(N\leq G\). The following are equivalent.
For all \(g\in G\), one has \(gN=Ng\). That is, \(N\trianglelefteq G\).
For all \(g\in G\), one has \(gNg^{-1} = N\).
For all \(g\in G\) and \(n\in N\), one has \(gng^{-1}\in N\).
Proof. WIP ◻
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