Advanced Algebra 2

This is a compiled set of notes separated by topic on advanced algebra, based on lectures at King's College London.

Group theory: Cosets and Quotient Groups

Definition: $H \subset G$ a subgroup of $G$. Translating from the left by some $g \in G$ gives a subset $gH = \{gh | h \in H\}$ which we call a left coset, also a corresponding right coset $Hg = \{hg | h \in H\}$.

Remark: H itself is a left $H$-coset: $H = eH$. Note also:

If $gH \cap g'H \neq \varnothing$, then $\exists k \in gH \cap g'H$ such that $gH = kH$ and $g'H = kH$ meaning $gH = g'H$. This means either $gH = g'H$ or $gH \cap g'H = \varnothing$.

There is a bijection from $H$ to any left $H$-coset $gH$ in $G$:

Example: $G = (\mathbb{Z},+)$, $H = \langle n \rangle$ for some $n >0$. For any $a \in \mathbb{Z}$ there is a coset $a+H = \{a +kn | k \in \mathbb{Z}\}$. Note $H = \langle n \rangle = \{kn | k \in \mathbb{Z}\}$.

Example: $G = S_3$, $H = \langle (123)\rangle = \{e,(123),(132)\}$ and $(12)H = \{(12),(23),(13)\}$. Since $|S_3| = 6$, these are the only two cosets.

Definition: We write $G/H$ for the set of all left $H$-cosets in $G$. We define the index of $H$ in $G$ to be $[G:H] = |G/H|$ which may be infinite.

Examples: $[\mathbb{Z}: \langle n \rangle] = |\mathbb{Z}/n \mathbb{Z}| = n$ $[S_3 : \langle (123) \rangle] = 2$ $[\mathbb{Z}: \{0\}] = \infty$.

Lagrange's Theorem: If $G$ is a finite group and $H$ a subgroup, then $|H| \vert |G|$, namely $|G| = |H|\cdot [G:H]$. Proof: $G$ is finite $\implies$ $H$ is finite, so $G/H$ is finite and $[G:H]$ is finite. Let $m \coloneq [G:H]$. Pick representatives $g_1,...,g_m$ so that:

are disjoint. Every element of $G$ lies in a coset, meaning:

as the cardinality of each $g_iH$ is $|H|$. $\blacksquare$

Corollary: For all $g \in G$, the order of $g$ divides $|G|$. Proof: Apply Lagrange's theorem to $H = \langle g \rangle$.

Definition: A subgroup $H$ of $G$ is called normal if:

We write $H \lhd G$ to say $H$ is a normal subgroup of $G$. We call $ghg^{-1}$ the conjugate of $h$ by $g$.

Definition: If $H$ is any subgroup of $G$ then $gHg^{-1}$ is again a subgroup and is called the conjugate subgroup of $H$ by $g$.

Lemma 1: A subgroup $H$ of $G$ is normal $\iff xHx^{-1} = H \forall x \in G$. Proof: $(\implies)$ If $H \lhd G \implies gHg^{-1} \subset H$. For $g=x$ we see $xHx^{-1} \subset H$, and likewise for $g=x^{-1}$ we get $x^{-1}Hx \subset H$, meaning $H \subset xHx^{-1}$ by multiplying by $x$ on the left and $x^{-1}$ on the right, and hence $H = xHx^{-1}$. The reverse direction of the proof is clear. $\blacksquare$

Lemma 2: Let $H$ be a subgroup of $G$. $H$ is a normal subgroup $\iff$ all left and right $H$-cosets agree. Proof: By Lemma 1: $H \lhd G \iff gHg^{-1} = H \ \forall g \in G$. Now notice $gHg^{-1} = H \iff gH = Hg$. $\blacksquare$

Examples:

• If $G$ is abelian then all its subgroups are normal.
• $G = S_3$: $H = \langle (123) \rangle$, then $eH = \{e, (123),(132)\} = He$ and $(12)H = \{(12),(23),(13)\} = H(12)$.
• $G = S_3$: $H = \langle (12)\rangle = \{e,(12)\}$ is not normal, i.e. $(13)(12)(13) = (23) \not\in H$.
• $G = GL_n(\mathbb{R})$ and $H = SL_n(\mathbb{R})$ this is a normal subgroup as $\det (AB) = \det(A) \det(B)$ for $n\times n$ matrices.

Quotient Groups

For a general subgroup $H \leq G$ we have that $G/H$ is just a set. If $H \lhd G$ then it turns out $G/H$ has a group structure.

Lemma: Let $H \lhd G$, the following defines a binary operation on $G/H$:

$(G/H,\star)$ is a group called the quotient group of $G$ by $H$.

Proof: To show $\star : G/H \times G/H \rightarrow G/H$ with $(gH,kH) \mapsto gH \star kH \coloneq gkH$ is well-defined. Suppose $g'H = gH$ and $k'H = kH$. Then:

• $g' = gh_1$ and $k'=kh_2$ for some $h_1,h_2 \in H$.
• $g'k' = gh_1kh_2 = gkk^{-1}h_1 k h_2 = gk (k^{-1}h_1 k)h_2 \in gkH$, since $H$ is normal. This means $g'k'H = gkH$, meaning $\star$ is well-defined. We can also check $(G/H,\star)$ is a group (Exercise).

Examples:

• $G = \mathbb{Z}$, $H = \langle n \rangle$ gives $\mathbb{Z}/\langle{n}$ as the integers modulo $n$.
• $H = \{rI_n | r \in \mathbb{R} - \{0\}\} \lhd GL_n(\mathbb{R})$ gives $PGL_n(\mathbb{R}) = GL_n(\mathbb{R})/H$

Homomorphisms and Isomorphisms

Let $(G,\star)$ and $(G',\star ')$ be groups. Definition: A function $f: G \rightarrow G'$ is called a group homomorphism if $f(g\star h) = f(g)\star 'f(h)$ for all $g,h \in G$.

Examples:

• $G = G' = \mathbb{R}$ with $\star = \star ' = +$. This means $f: \mathbb{R} \rightarrow \mathbb{R}$ where $x \mapsto 2x$ is a homomorphism, but $g: \mathbb{R} \rightarrow \mathbb{R}$ where $x \mapsto x^2$ is not a homomorphism.
• $G = G' = \mathbb{R}^{\times} = \mathbb{R} - \{0\}$ with $\star = \star ' = \cdot$. This means $f: \mathbb{R}^{\times} \rightarrow \mathbb{R}^{\times}$ where $x \mapsto 2x$ is not a homomorphism but $g: \mathbb{R}^{\times} \rightarrow \mathbb{R}^{\times}$ where $x \mapsto x^2$ is a homomorphism.
• $G = GL_n(\mathbb{R})$ and $f: G \rightarrow G$ where $A \mapsto A^2$ is not a homomorphism if $n\geq 1$, however $f:G\rightarrow \mathbb{R}^{\times}$ where $A \mapsto \det(A)$ is a homomorphism.

Remarks:

• If $f: G \rightarrow G'$ is a group hom. then:
  • $f(e) = e'$
  • $f(g^{-1}) = f(g)^{-1}$
  • $f(g^n) = f(g)^n$ for $n \in \mathbb{Z}$.
• A composite of homomorphisms is a homomorphism:
  • $f,f'$ homs: $G \xrightarrow{f}G' \xrightarrow{f'}G'' \implies f'\circ f$ is a hom.

Definition: $f: G \rightarrow G'$ a group hom $\implies \ im(f) = f(G) \subset G'$ is a subgroup, and $\ker(f) = \{g \in G | f(g) = e'\} \subset G$ is a subgroup.

Example:

• Suppose $N \lhd G$, then $f: G \rightarrow G/N$, $g \mapsto gN$ is a homomorphism.
• $\ker(f) = N$.

Remark:

• $f$ is surjective $\iff \ im(f) = G'$.
• $f$ is injective $\iff \ \ker(f) = \{e\}$.

Definition:

• A bijective hom $f: G \rightarrow G'$ is called an isomorphism, denoted $f: G \xrightarrow{\sim} G'$.
• A bijective hom $f: G \rightarrow G$ is called an automorphism, denoted $f: G \xrightarrow{\sim} G$. Write $G \cong G'$ and call $G, G'$ isomorphic if $\exists \ G \xrightarrow{\sim} G'$. "$\cong$" is an equivalence relation on the set of groups.

Examples:

• $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto 2x$ is an automorphism of $(\mathbb{R},+)$.
• $f: (\mathbb{R},+) \rightarrow (\mathbb{R}_{>0},\cdot)$, $x \mapsto e^x$ is an isomorphism.
• $f: \mathbb{R} \rightarrow \left\{\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} | x \in \mathbb{R} \right\}$, $x \mapsto \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$is an isomorphism of groups.

First Isomorphism Theorem: Let $f: G \rightarrow G'$ be a group hom, and $N = \ker(f)$. Then:

1. $N \lhd G$
2. $\overline{f} : G/N \rightarrow G'$, where $gN \mapsto f(g)$ is a well-defined injective group hom.
3. $G/N \xrightarrow{\sim} im(f)$ with isomorphism $gN \mapsto f(g)$.

Proof:

1. $N = \ker(f) = \{ n \in G | f(n) = e'\}$, pick any $g \in G$ and $n \in N$. Then $f(gng^{-1}) = f(g)f(n)f(g)^{-1} = f(g)f(g)^{-1}= e'$. This implies $gng^{-1} \in \ker(f) = N \implies N \lhd G$.
2. Note $\overline{f}: G/N \rightarrow G'$ where $gN \mapsto f(g)$ is a well-defined injective group hom. Assume $gN = g'N \implies g = g'n$ for some $n \in N$. This means $f(g) = f(g'n) = f(g')f(n) = f(g')e' = f(g')$ This means $\overline{f}(gN) = \overline{f}(g'N)$. $\ker (\overline{f}) = \{eN\}$ which is the identity element of $G/N$, meaning $\overline{f}$ is injective.
3. $G/N \xrightarrow{\sim} im(f)$ with $gN \mapsto f(g)$ is an injective group hom by (2), and it is surjective by construction since $im(f) = im(\overline{f})$. $\blacksquare$

We may use the notation $\overline{g} = gN$ so that $\overline{f}(\overline{g}) = f(g)$.