Groups

0 - Magma, Semigroups, Monoids, Groups

Definition (Magma): A set $S$ with a map $S\times S \overset{\circ}{\longrightarrow} S$ is called a magma. The map $\circ$ is called a binary operator (or just operator), and for $(x,y) \in S\times S$, we write $\circ (x,y)$ as $x \circ y$ , and often shorten it to just $xy$. Note that $xy$ is called the product of $x$ and $y$.

A binary operator of a magma is associative if $x(yz) = (xy)z$. In this case we can use induction to show that the product of $n$ elements can simply be written $x_1x_2,\dots x_{n}$ , and any parenthesizing of this product is equivalent.

A binary operator of a magma is commutative if $xy = yx$. It is customary to use additive notation in such cases, that is we often denote $\circ (x,y)$ as $x+y$, when $\circ$ is commutative.

Definition (Identity element): let $S$ be a magma. an element $e\in S$ is called the identity element (or just identity) if for every $s \in S$, $se = es = s$. Note that the identity of a magma is unique. Otherwise, suppose $e'$ is also an identity of $S$, then $e'e = e = e'$.

Definition (Semigroup): A semigroup $S$ is a magma whose operator is associative.

Definition (Monoid): A monoid $S$ is a semigroup with a (unqiue) identity.

Definition (Group): A monoid $G$ where each element $g \in G$ has an inverse $g^{-1} \in G$ such that $gg^{-1} = g^{-1}g = e$ is a group. A group is abelian if it’s operator is commutative.

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Proposition (Uniqueness of inverses): Let $G$ be a group, then for each $g \in G$, there is a unique $g^{-1} \in G$ such that $gg^{-1} = g^{-1}g = e$.

Proof: Suppose both $g_{}^{-1}, g_{-}^{-1}$ are inverses of an arbitrary element $g \in G$. Then, $e = gg^{-1}$ multiplying $g_{-}^{-1}$ on boh sides, we get $g_{-}^{-1} e = (g_{-}^{-1}g)g^{-1}$. Therefore, $g_{-}^{-1} = g^{-1}$. $\square$

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Proposition (The unique property of the identity): let $G$ be a group, then the identity $e \in G$ is the only element in $G$ such that $e^2 = e$

Proof: Suppose for some $g\in G$, we have that $gg = g$. Then, multiplying by $g^{-1}$ on both sides, we have $g = e$. $\square$

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Proposition (A one way condition to guarantee abelian groups): let $G$ be a group, such that for each $g \in G$, $g^2 = e$, then $G$ is abelian.

Proof: let $x,y$ be arbitrary elements in $G$. Then, $xy \in G$. Hence there is a unique $(xy)^{-1} \in G$, such that $(xy)(xy)^{-1} = e$. Using the algebraic properties of groups, it is not hard to see that $(xy)^{-1} = y^{-1}x^{-1}$. However, its given to us that $(xy)(xy) = e$. which means that $(xy) = (xy)^{-1}$ (they are “both” the unique inverse elements of $xy$). Equivalently, $xy = y^{-1}x^{-1}$. Again, in $G$, $x x = e = y y$. Hence, $x = x^{-1}$ and $y = y^{-1}$. Therefore, $xy = yx$, hence $G$ is abelian.

To show that this condition is only one way, consider the group of integers with addition as the operator, $\mathbb Z, +$. This group is abelian, however, it is not true that $x + x = 0$ for each $x \in \mathbb{Z}$. $\square$

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Proposition (Left identity + left inverses suffice): let $G$ be a semigroup, such that there exists $e\in G$ such that $eg = g$ for all $g\in G$, and for each $g \in G$, there exists $g^{-1} \in G$ such that $g^{-1} g = e$. Then $G$ is a group.

Proof: we play some algebraic tricks here: Let $g \in G$, then $eg = g$. Consider the equation $ge = ege$. Notice that $g^{-1}g = e = {g^{-1}}^{-1} g^{-1}$ (every element has a left inverse). Now, $ge = ({g^{-1}}^{-1} g^{-1}) \ (g) \ (g^{-1}g)$. Hence, upon simplifying, we have $ge = g$. Hence $e$ is also a right identity, therefore, $e$ is the identity. Similarly, $gg^{-1} = egg^{-1}$. Hence, $gg^{-1} = ({g^{-1}}^{-1} g^{-1}) gg^{-1}$. Upon simplifying, we have $gg^{-1} = e$. Therefore $g^{-1}$ is also the right inverse and hence the inverse of $g$. $\square$

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Proposition (Right identity + right inverses suffice): Give a proof, symmetrical to the one above.

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Proposition (Finite semigroup with cancellation is a group): Let $G$ be a finite semigroup, $|G| = n$ , such that $ab = ac$ implies $b = c$ and $ba = ca$ implies $b = c$. Then $G$ is a group.

Proof: for an arbitrary element $g \in G$, consider the map $G \overset{f_{g}}{\longrightarrow} G$, given by $f_g(x) = gx$ for all $x \in G$. This map is injective, as $gx_1 = gx_2$ implies $x_1 = x_2$. Hence, each $x \in G$ goes to a unique $gx \in G$. therefore, the image of $G$ under $f_g$ has exactly $n$ elements, Hence $fg(G) = G$, therefore, the map is bijective. hence, there exists a unqiue $e \in G$ such that $f_g(e) = g$ that is $ge = g$. Now consider an arbitrary element $h \in G$. notice that $gh = (ge)h$. Hence $h = eh$. therefore $e$ is the left identity. moreover, $hg = h(eg)$ (e is the left identity) Hence, $h = he$ therefore, $e$ is also the right identity. Now, for an arbitrary element $g$ the map $f_g$ as described above is bijective, so there is a unique $g^{-1}$ such that $f_g(g^{-1}) = e$. Equivalently every element $g$ has a unique right inverse $g^{-1}$ such that $gg^{-1} = e$. We finish the proof by applying the proposition on right identity + right inverses. $\square$

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Product of Groups

Definition (Direct product of groups): Let $G_1, G_2,\dots G_{n}$ be a collection of groups. Then, the set $\prod_{i=1}^{n} G_{i}$ with the product of two tuples $\boldsymbol g^* = (g_{1},g_{2}, \dots g_{n})$ and $\boldsymbol h^* = (h_1, h_{2}, \dots h_{n})$, where $g_i, h_i \in G_i$. Is given by $\boldsymbol {g^*h^*} = (g_{1}h_{1}, g_{2}h_{2}, \dots g_{n} h_{n})$ is called the direct product of a collection of groups.

Proposition (The direct product of groups is a group):

Proof (outline): Associativity of the product defined above, follows directly from the associativity of the products of each group $G_i$. The tuple of identity elements from each group $\boldsymbol{e^*} = (e_1, e_{2}, \dots e_{n})$ acts as the identity element, and for the arbitrary element $\boldsymbol g^* = (g_{1},g_{2}, \dots g_{n})$, the inverse element is given by $\boldsymbol{{g^*}^{-1}} = (g^{-1}_{1},g^{-1}_{2}, \dots g^{-1}_{n})$. $\square$

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1 - Maps on Groups

Definition (Group homomorphism): let $G_1, G_2$ be groups and $f: G_1 \rightarrow G_2$ be a map. $f$ is a group homomorphism if for each $x,y \in G_1$, $f(x \cdot_{G_1} y) = f(x) \cdot_{G_2} f(y)$.

furthermore, we will suppress any notation to indicate that $xy$ is a product in $G_1$ and $f(x)(fy)$ is a product in $G_2$.

Proposition (Properties of homomorphisms): For the following discussion, let $G_1, G_2$ be groups, with identities $e_1$ and $e_2$ and $f$ a homomorphism from $G_1 \ to \ G_2$.

1. $f(e_1) = e_2$ (under a homomorphism, the image of the identity of the domain group, is the identity of the target group).
2. for each $x \in G_1$, $f(x)^{-1} \in G_2$ is $f(x^{-1})$ where $x^{-1}$ is the inverse $x$ in $G_1$.
3. Let $G_1, G_2, G_3$ all be groups, and let $f$ be a homomorphism from $G_1$ to $G_2$, $h$ a homomorphism from $G_2$ to $G_3$, then $h\circ f$ is a homomorphism from $G_1$ to $G_3$.(the composition of homomorphisms, is itself a homomorphism).

Proof:

1. We will use the unique property of the identity element of a group. Notice that $f(e_1e_1) = f(e_1)f(e_1) = f(e_{1})$. Hence it must be that $f(e_1) = e_2$.
2. Notice that for any $x\in G_1$, $f(xx^{-1}) = f(x)f(x^{-1})$ but $xx^{-1} = e_{1}$. Therefore, $e_2 = f(x)f(x^{-1})$ Hence $f(x^{-1})$ is the unique inverse of $f(x)$ in $G_2$. In notation, $f(x)^{-1} = f(x^{-1})$.
3. let $x,y$ be arbitrary elements of $G_1$. Then, $h(f(xy)) = h(f(x)f(y)) = h(f(x)) \ h(f(y))$. $\square$

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Definition (Group isomorphism): let $G_1$, $G_2$ be groups. a map $f: G_1 \rightarrow G_2$ is an isomorphism, if it is a bijective homomorphism. An isomorphism from $G$ onto itself is called an automorphism.

Proposition (The inverse of an isomorphism is an isomorphism): if $f: G_1 \rightarrow G_2$ is an isomorphism, then $f^{-1} : G_2 \rightarrow G_1$ is an isomorphism, where $G_1, G_2$ are groups.

Proof: The inverse of a bijection, is also a bijection. Hence, it is sufficient to prove that $f^{-1}$ is a homomorphism. let $u,v$ be arbitrary elements of $G_2$. Then, since $f$ is bijective, there exists unique $x,y \in G_1$ such that $u = f(x), v = f(y)$. hence $uv = f(x)f(y) = f(xy)$. Now, $f^{-1}(uv) = f^{-1}f(xy) = xy = f^{-1}(u) f^{-1}(v)$. Hence $f^{-1}$ is a homomorphism. $\square$

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Proposition (Automorphisms form a group): Let $G$ be a group, and let $\text{Aut}(G)$ be the collection of all automorphisms on $G$. Then this set, with map composition as the operator, forms a group.

Proof: notice that the identity map $id(x) = x$ from $G$ to $G$ is an automorphism (it is both bijective and homomorphic). And acts as the group identity under composition. for any two maps $f,h \in \text{Aut}(G)$. Then $hf$ is bijective, moreover by the properties of homomorphisms, $hf$ is a homomorphism, hence $hf \in \text{Aut}(G)$. Associativity follows as a property of function composition, and since the inverse of an isomorphism is an isomorphism, for each $f \in \text{Aut}(G)$, $f^{-1} \in \text{Aut}(G)$ : the inverse map, acts as the group inverse element. $\square$

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The Left Coset Map

Definition (Left coset maps): Let $G$ be a group, and $g$ an arbitrary element of $G$. then the left coset map $f_g : G \rightarrow G$ is the map $f_g(x) = gx$ for each $x\in G$.

Proposition (Left coset maps are bijective): Let $G$ be a group, and $L(G)$ be the collection of left coset maps from $G$ to $G$. Then, every element $f_g \in L(G)$ is a bijection on $G$.

Proof: Let $g$ be an arbitrary element of $G$. We claim that $f_g$ is a bijection. suppose for $x,y \in G$, $f_g(x) = f_g(y)$. Then, $gx = gy$ hence $x = y$, showing injectivity. Now, suppose $y$ is an arbitrary element in $G$. Notice that $f_g(g^{-1}y) = g(g^{-1}y) = y$. Showing Surjectivity. $\square$

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Theorem (Left coset maps form a group isomorphic to $G$): Let $G$ be a group, then the collection of left coset maps $L(G)$ is a group under map composition. Moreover, $L(G)$ is isomorphic to $G$.

Proof: The identity map on $G$ is $f_e = ex$ for all $x\in G$, $f_e \in L(G)$ is the group identity as $f_e f_g = f_g f_e = f_g$ for any $f_g \in L(G)$. Associativity follows as a property of map composition, of course, the composition of two left coset maps $f_g \ , f_h \in L(G)$ . is $f_g f_h = f_{gh}$ , because for all $x \in G$ $f_g(f_h(x)) = g(hx) = (gh)x =f_{gh}(x)$. Hence, $L(G)$ is closed under composition, moreover the map $g \mapsto f_g$ is a homomorphism, as $gh \mapsto f_{gh} = f_g f_h$ . For each $f_g \in L(G)$ the unique $f_g^{-1}$ is given by $f_{g^{-1}}$. This is because for each $x\in G$, $f_g(f_{g^{-1}} (x)) = f_{g^{-1}}(f_{g}(x)) = gg^{-1}x = g^{-1}gx = ex$. Hence $f_gf_{g^{-1}} = f_{g^{-1}}f_g = f_e$. We claim that the map $g \mapsto f_g$ is bijective. Surjectivity is obvious. now suppose that $f_{g_1} = f_{g_2}$ then for each $x\in G$, $g_1x = g_2x$ hence $g_1 = g_2$. Hence we produce the desired isomorphism from $G$ to $L(G$). $\square$

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Lemma (Distinct left coset maps never agree): Let $G$ be a group, and $f_g, f_h \in L(G)$ be two distinct left coset maps. Then there is no $x \in G$ such that $f_g(x) = f_h(x)$. Otherwise, this would imply that $gx = hx$ hence $g = h$ (multiplying $x^{-1}$ on both the sides), meaning that $f_g$ and $f_h$ are identical maps.

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Warning: A left coset map is NOT a homomorphism. Let $G$ be a group. Then a map $f_g \in L(G)$ if $g \ne e$ then $f_g$ is not a homomorphism from $G$ to $G$ . $f_g(xy) = g(xy)$ , $f_g(x)f_g(y) = gxgy$. Even if $G$ is abelian, $gxgy = g^2xy$ and $g^2 = g$ only if $g = e$. See the unique property of the identity of a group.

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2 - Subgroups

Definition (Subgroup): let $G$ be a group, then a subset $H$ of $G$ is a subgroup of $G$ if it a group, inheriting the operator from $G$. In such a case we write $H \leqslant G$. if $H \ne G$, we write $H < G$.

Proposition (Necessary and sufficient conditions for subgroups): let $G$ be a group, $H$, a subset of $G$ is a subgroup if and only if the following properties hold:

1. The identity $e \in G$ is also in $H$.
2. $x,y \in H$ implies that $xy \in H$ ($H$ is closed under the product operator inherited from $G$).
3. $x \in H$ implies that $x^{-1} \in H$ ($H$ is closed under inverses).

Proof: if $H$ is a subgroup, 1,2,3 follow by definition. Now suppose $H \subset G$. Then the above three conditions guarantee that $H$ is a group, with associativity following from inheriting the operator from $G$. $\square$

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Theorem (A simpler subgroup condition): let $G$ be a group, and $H$ a subset of $G$. then $H$ is a subgroup if and only if $H$ is non empty, and $x,y \in H$ implies $xy^{-1} \in H$.

Proof: Suppose $H$ was a subgroup of $G$. then obviously, $H$ is non empty (it at least contains $e$, the identity) and $x,y \in H$ implies $xy^{-1} \in H$. Now, suppose $H$ was a non empty subset of $G$, and that $x,y \in H$ implies $xy^{-1} \in H$. We will show that $e \in H$: notice that $x \in H$ implies $xx^{-1} = e \in H$. Now we show that $H$ is closed under inverses: Since $e \in H$, for any element $x \in H$, $ex^{-1} = x^{-1} \in H$. Finally, we will show that $H$ is closed under products: Suppose $x,y \in H$. Then $y^{-1} \in H$, (as shown earlier). Hence $x(y^{-1})^{-1} \in H$. Hence $xy \in H$. Therefore by the necessary and sufficient conditions above, $H$ is a subgroup of $G$. $\square$

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Proposition (Intersection of subgroups): Let $G$ be a group, and $H, K$ be subgroups of $G$. Then, $H \cap K$ is a subgroup of $G$.

Proof: since $e$ is in both $H$ and $K$, $H \cap K$ is non empty. Suppose $x,y \in H \cap K$ . Then $x,y$ is in both $H$ and $K$. Therefore, $xy^{-1}$ is in both $H$ and $K$. Hence, $xy^{-1} \in H \cap K$. Apply the simpler subgroup condition. $\square$

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Theorem (Intersection of finitely many subgroups): Let $H_1, H_2,\dots H_{n}$ All be subgroups of a group $G$. Then, $\bigcap_{i = 1}^{n} H_{i}$ is a subgroup of $G$.

Proof outline: Use the intersection of subgroups proposition with a natural induction argument.

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Theorem (Finiteness + product closure suffice for subgroups): Let $H$ be finite subset of a group $G$, such that the identity $e \in H$. If $x,y \in H$ implies $xy \in H$, then $H$ is a subgroup of $G$.

Intuition:

Imagine a finite set $H$, closed under products, with the identity element $e$ in it. Starting from an arbitrary element $x$, start taking left products, $h_1x$, $h_2h_1 x$ and so on… each of these products are the images of $x$ under distinct left coset maps. so eventually, we have to hit $e$, (we will show that we can’t loop around without ever hitting $e$), basically allowing us to get closure under inverses.

Proof: Since $H$ is finite, let $|H| = n$, we can index the elements of $H =\{ h_{0} = e, h_{1}, \dots h_{n-1} \}$. We claim that the left coset maps $L(H)$ are well defined from $H$ to $H$. suppose $x \in H$ and $h_i \in H$. then $f_{h_i}(x) = h_ix \in H$ due to product closure. Now, Consider the set of orbits of $x \in H$ under each map in $L(H)$ defined as $Orb(x) = \{h_ix : h_i \in H\}$. By the lemma on distinct left coset maps, if $h_ix = h_jx$, then in $G$ which has inverses, $h_i = h_j$ meaning that they are the identical left coset map in $G$, Hence, they are identical in $H$. Therefore $Orb(x)$ has $n$ distinct elements, each of which lie in $H$. Hence $Orb(x) = H$, given that $H$ is finite. Since $Orb(x) = H$ there exists some $h_j \in H$ such that $h_j x = h_0 = e$. This is the inverse of an element $x$. Hence $H$ is closed under inverses. Apply the necessary and sufficient conditions for subgroups. $\square$

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3 - Fibers, Kernel, Conjugation, Centre

Definition (Fibers, Images): note: This term is inherited from category theory, where it carries more meaning. But in our case, it’s just a fancy word for the inverse image of set over a map.

let $G_1, G_2$ be groups and $f: G_1 \rightarrow G_2$ be a map, then the fibers of $H_2 \subset G_2$ over $f$ is the set $f^{-1}(H_2) =\{ x \in G_{1} : f(x) \in H_{2} \}$.

The image of a set $H_1 \subset G_1$ is the set $f(H_1) =\{ f(x) : x \in H_{1}\}$.

Definition (Kernel): let $G_1, G_2$ be groups and $f: G_1 \rightarrow G_2$ be a map. Then the kernel of $f$, $Ker(f)$ is the fibres of the identity $e_2 \in G_2$. that is, $Ker(f) = f^{-1}(\{e_2\}) =\{ x \in G_{1}: f(x) = e_{2} \}$.

Definition (Conjugation, Normalizer): let $G$ be a group, and $S$ a subset of $G$. Let $g$ be any particular element of $G$. Then the conjugation of $S$ with $g$ is the set $gSg^{-1} =\{ gsg^{-1} : s \in S \}$.

The normalizer of $S$, $N(S)$ is the set of all $g \in G$ for which the conjugation of $S$ with $g$ is equal to $S$ itself. That is, $N(S) = \{ g \in G : gSg^{-1} = S \}$.

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Proposition (Kernel is a subgroup): Let $G_1, G_2$ be groups and $f: G_1 \rightarrow G_2$ a homomorphism. Then, $Ker(f) \leqslant G_1$.

Proof: We will use properties of homomorphisms for the following argument. Notice that $f(e_1) = e_2$ hence $e_1 \in Ker(f)$.

Now suppose $x,y \in Ker(f)$ then $f(x) = f(y) = e_2$. Hence $f(x)f(y) = e_2e_2 = e_2$ (see: the unique property of the identity element of a group). But, $f$ is a homomorphism. Hence, $f(xy) = e_2$. Therefore, $xy \in Ker(f)$.

Now, suppose $x \in Ker(f)$, then consider the equation $f(xx^{-1}) = f(e_1) = e_2$. Since $f$ is a homomorphism, we obtain $f(x)f(x^{-1}) = e_2$. But, $f(x) = e_2$. Therefore, $f(x^{-1}) = e_2$ hence $x^{-1} \in Ker(f)$ . $\square$

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Proposition (Image is a subgroup): Let $G_1, G_2$ be groups and $f: G_1 \rightarrow G_2$ a homomorphism. Then, $f(G_1) \leqslant G_2$.

Proof: First, notice that $f(G_1)$ is non empty, as $e_1 \in G_1$, hence $e_2 \in f(G_1)$. Now, suppose $f(x),f(y) \in f(G_1)$ . Then, $f(x)f(y) = f(xy) \in f(G_1)$ as $xy \in G_1$. Similarly, if $f(x) \in G_1$, then $f(x)^{-1} = f(x^{-1}) \in f(G_1)$ as $x^{-1} \in G_1$. See: properties of homomorphisms. $\square$

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Proposition (Conjugation of a subgroup): Let $G$ be a group, and $H$ be a subgroup of $G$. then for any element $g \in G$, $gHg^{-1} \leqslant G$.

Proof: Since $e \in H$ , $geg^{-1} = e \in gHg^{-1}$.

Now, suppose $u,v \in gHg^{-1}$. Then for some $x,y \in H$, $u = gxg^{-1}$ and $v = gyg^{-1}$. Therefore, $uv = g(xy)g^{-1}$, hence $uv \in gHg^{-1}$.

Now, suppose $t \in gHg^{-1}$ then $t = ghg^{-1}$ for some $h \in H$. $t^{-1} = gh^{-1}g^{-1}$ (The reader is asked to pause and carry out the product $tt^{-1}$ and $t^{-1}t$, should any doubt arise). Finally apply the necessary and sufficient conditions for subgroups. $\square$

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Proposition (Normalizer of a subgroup): Let $G$ be a group, and $H$ a subgroup of $G$, then, $H \leqslant N(H) \leqslant G$.

Proof: We will begin by showing that $N(H)$ is a subgroup of $G$:

$N(H) =\{ g \in G : gHg^{-1} = H \}$. Clearly, $N(H)$ is non empty as $eHe = H$. ($e = e^{-1}$).

Suppose $g,h \in N(H)$. then $gHg^{-1} = hHh^{-1} = H$. Notice that $(gh)H(h^{-1}g^{-1}) = g(hHh^{-1})g^{-1} = g(H)g^{-1} = H$. Therefore, $gh \in N(H)$.

Now suppose that $g \in N(H)$, then $gHg^{-1} = H$. But $eHe = H$. Hence, $g^{-1}(gHg^{-1})g = H$, therefore, $g^{-1}Hg = H$. Hence $g^{-1} \in N(H)$.

Finally, apply the necessary and sufficient conditions for subgroups.

Now, we will show that $H$ is a subgroup of $N(H)$.

Of course $H$ is a group, so it is sufficient to show that $H \subset N(H)$. For any $h \in H$, $hHh^{-1} = (hH)h^{-1}$ but $hH$ is just the image of the left coset map on $H$ given by $f_h(x) = hx$ for all $x \in H$. That is, $hH$ is just notation for $f_h(H)$. But left coset maps are bijective. Hence, $hH = H$.

Therefore, we have $hHh^{-1} = (hH)h^{-1} = Hh^{-1}$. Using a similar argument, $Hh^{-1} = g_{h^{-1}}(H)$ Where $g_{h^{-1}}$ is the right coset map induced by $h^{-1}$. Analogously, right coset maps are bijective too. hence $Hh^{-1} = H$.

Finially, this gives us $hHh^{-1} = H$. Therefore, any $h\in H$ is also in $N(H)$. $\square$

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Definition (Center of a group): The Center of a group $G$, denoted by $Z(G)$, is the subset of $G$ whose all elements commute with the entirety of $G$. That is $Z(G) =\{ g\in G : gx = xg, \ \forall x \in G \}$.

Proposition (The center is a subgroup): Let $G$ be a group. Then, $Z(G) \leqslant G$.

Proof: Since $e \in Z(G)$, $Z(G)$ is not empty. Let $g,h \in Z(G)$. then for all $x\in G$, $gx = xg$ and $hx = xh$. But, $hx, xh \in G$. Hence $(gh)x = (gx)h = x(gh)$ Hence $gh \in Z(G)$. Now, if $x \in G$, and $g \in Z(G)$, then $x^{-1} \in G$. Hence, $gx^{-1} = x^{-1}g$ hence $xg^{-1} = g^{-1}x$. hence $g^{-1} \in Z(G)$. $\square$

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4 - Examples of Groups

The Klein-Four Group

A way to generate the Klein-Four Group: Notice that except the identity permutation $k_0$ , every other permutation, can be written as the composition of two independent swaps. Let $(i,j)$ denote the permutation $i \mapsto j, \ j \mapsto i$ And leaves everything unchanged. Then, $k_1 = (1,3)(2,4)$ , $k_2 = (1,2)(3,4)$ and, $k_3 = (1,4)(2,3)$. Furthermore, notice that $k_1^2 = k_2^2 = k_3^2 = k_0$. and $k_1k_2 = k_2k_1 = k_3$ and so on…

Definition (Klein-Four group): The set $K_4 = \{e, a, b, c\}$ with $a^2 = b^2 = c^2 = e$ and $ab = ba = c$, $ac=ca = b$, $bc= cb = a$ as the products is called the Klein-Four group.

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The Dihedral Group

A necklace for a group: Below, we depict two necklaces, one on a rigid triangular wire, and one on a rigid square wire. We can imagine distinctly colored beads on each vertex. The question then is, how can we manipulate the necklace, so that when you place it on the table and take a picture, it looks different in the arrangement of beads? call the “do nothing” transformation $e = r_{0}$. Then, call the counter-clockwise rotation by $2\pi/n$ (for a regular $n$-gon), $r_{1}$. Then every rotation of the necklace is given by $r_0, r_1,\dots , r_{n-1}$. Where $r_n = r_0$.

Now notice that you’ve got a “front” of the necklace and “back” of the necklace. And to get to the back of the necklace from the front, you would reflect it about an axis, say $S_0$ call that $s_0$. notice that two reflections about the same axis is equivalent to “doing nothing”. $s_i^2 = r_0$.

And by observing this picture more, one can work out all the other products. Two reflections should be a rotation, because you come back to the “front” of the necklace, and so on…

A rotation, then a reflection leaves you in the “back” of the necklace, so it should be equivalent to a reflection about some axis…

Definition (Dihedral group of a regular $n$-gon): The set $D_n=\{r_{0},r_{1},\dots r_{n-1},s_{0},s_{1},\dots s_{n-1} \}$ with the following products:

• $r_{i}r_{j} = r_{i+1}$
• $s_is_{j} = r_{i-j}$
• $s_{i}r_{j} = s_{i-j}$
• $r_{i}s_{j} = s_{i+j}$

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The Additive and Multiplicative Groups on the Reals, The Circle Group

Definition (Additive and multiplicative groups on Reals): The set $\mathbb R$ with addition $+$ forms an abelian Group. The set $\mathbb R - {0}$ with multiplication $\cdot$ forms an abelian group as well.

The map $x \mapsto e^{x}$ is the desired isomorphism from $(\mathbb R , +)$ to $(\mathbb R - {0}, \cdot)$

Definition (The circle group): In the complex numbers $\mathbb C$, the unit circle $S^1 =\{ e^{i\theta}: \theta \in \mathbb R \}$ is an abelian group with complex multiplication.

$\theta \mapsto e^{i\theta}$ is the desired homomorphism from $(\mathbb R , +)$ to $S^1$. if $\theta$ is restricted to $[0,2\pi)$, then the given map is an isomorphism.

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5 - Generators, Cyclic Groups

Definition (Containment of a subset): Let $G$ be a group and $S$ be a subset of $G$. Then a containment group of $S$, denoted by $\bar S$ is the collection of subgroups $H$ of $G$ such that $S \subset H$. That is $\bar S = \{ H \leqslant G : S \subset H\}$.

Now, Notice that any $H \in \bar S$ has the following properties:

1. Since $H$ contains $S$, for each $s \in S$, $s^{-1} \in H$.
2. For any $m \in \mathbb N$, and $s_1,s_{2}, \dots s_{m} \in S$, the product $s_1s_2...s_m \in H$.
3. Putting it all together, for any $m \in \mathbb N$, $H$ contains the element $s_1s_2\dots s_{m}$ where $s_i \in S$ or $s_i^{-1} \in S$.

Now, define $\langle S \rangle=\{ s_{1}s_{2}\dots s_{m} : m\in \mathbb N, s_{i} \in S \ or \ s_{i}^{-1} \in S \}$ . From the above discussion, it is clear that $\langle S \rangle \subset H$ for each $H \in \bar S$. Moreover, of course $S \subset \langle S \rangle$. We show that $\langle S \rangle$ is a group (precisely it is a subgroup of $G$).

We will repeatedly abuse the definition of $\langle S \rangle$ (note that a definition is not a person so this is okay) for the following proof.

1. $e \in \langle S \rangle$ : pick any $s \in S$. Then, by defenition, it is clear that $ss^{-1} = e \in \langle S \rangle$.
2. Suppose $x,y \in \langle S \rangle$ Then $x = s_1s_2\dots s_{m}$ and $y = t_1t_2\dots t_{n}$ for some $m,n \in \mathbb N$, such that $t_i \ or \ t_i^{-1} \in S$ and $s_i \ or \ s_i^{-1} \in S$. Hence, $xy = s_{1}s_{2}\dots s_{m}t_{1}t_{2}\dots t_{n} \in \langle S \rangle$.
3. Suppose $x \in \langle S \rangle$ Then $x = s_1s_2\dots s_{m}$, for some $m \in \mathbb N$ such that $s_i \ or \ s_i^{-1} \in S$. Then $x^{-1} =s_{m}^{-1}s_{m-1}^{-1} \dots s_{1}^{-1} \in \langle S \rangle$.

Hence, each $H \in \bar S$ contains the subgroup $\langle S \rangle$. Therefore, $\langle S \rangle$ is the smallest group that contains $S$, known as the subgroup generated by $S$. In particular, for each $H \in \bar S$, $S \subset \langle S \rangle \leqslant H$.

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Definition (Group generated by a set): let $G$ be a group, and $S$ a subset of $G$, then inheriting the operator from $G$, the subgroup of $G$ generated by $S$, is the set $\langle S \rangle =\{ s_{1}s_{2}\dots s_{m} : m\in \mathbb N, s_{i} \in S \ or \ s_{i}^{-1} \in S \}$.

Definition (Cyclic group): For a group $G$, the group generated by any element $g \in G$ is denoted by $\langle g \rangle$. This group is called the cyclic subgroup of $G$ generated by $g$.

Notice that $\langle g \rangle$ just contains the products of $g$ with itself : $g^i$ and the products of $g^{-1}$ with itself: $g^{-i}$, where $g^0 := e$. The smallest $n \in\mathbb N$ for which $g^n = e$ is (if it exists) known as the order of $\langle g \rangle$.

In such a case, it is clear that $\langle g \rangle =\{ g^0 = e, g^1, \dots g^{n-1}\}$. where ${g^i}^{-1} = g^{n-i}$.

Otherwise, if no such $n$ exists, the order of $\langle g \rangle$ is infinite.

The order of an element $g$ of a group $G$ is the order of its cyclic subgroup $\langle g \rangle$.

Two cyclic groups of the same order $\langle g \rangle$, $\langle h \rangle$ are isomorphic to each other, where $g^{i} \mapsto h^{i}$ is the desired isomorphism.

Moreover, a cyclic group of finite order, say $n$ is isomorphic to the rotations of a regular $n$-gon.

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Proposition (Necessary and sufficient condition for a cyclic group): Let $G$ be a group. Then $G$ is cyclic if and only if, there exists $g \in G$, $G = \langle g \rangle$.

Proof: Suppose $G$ is cyclic, then by definition there exists an element $g \in G$ such that $G =\{ {g}^{i} : i \in \mathbb Z \}$. Now suppose that there exists $g \in G$ such that $G = \langle g \rangle$, then by definition, $G$ is cyclic. $\square$

Although the above proposition seems to bash on the obvious, it’s extremely useful: for example in showing that the multiplicative group of integers, modulo a prime is cyclic.

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6 - Symmetric Groups

Definition (Symmetric group): Let $S$ be a set, then the set of all bijection on $S$, with function composition as the operator, is a group, called the symmetric group on $S$, denoted by $Sym(S)$.

Here, the identity map $id: S \rightarrow S$ acts as the group identity, with any $f \in Sym(S)$ $id \ f = f \ id = f$.

The inverse function $f^{-1}$ acts as the inverse element of each $f \in Sym(S)$.

Associativity and closure follow as properties of composition of bijections.

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Theorem (Set cardinalities categorize symmetric groups): Let $S$ and $T$ be sets, such that the cardinalities of $S$ and $T$ are equal. Then $Sym(S)$ is isomorphic to $Sym(T)$

Commutative Diagram:

Proof: Let $S$, $T$ be sets with equal cardinality, and $\phi : S \rightarrow T$ be a bijection. Moreover, let $f_S$ denote a bijection on $S$, and $f_T$ denote a bijection on $T$.

By observing the commutative diagram above, we see that for every $f_T \in SYM(T)$ there exists an $f_S$ such that $f_S = \phi^{-1}f_T\phi$ . Note that this indeed a bijection on $S$, because the composition of bijections is a bijection.

Hence we claim that the map $f_T \mapsto \phi^{-1}f_T\phi$ is an isomorphism from $SYM(T)$ to $SYM(S)$.

Homomorphicity: $F_{T1} \ F_{T2} \mapsto \phi \ F_{T1} \ F_{T2} \ \phi^{-1} = (\phi^{-1}\ f_{T_1} \ f_{T_2} \ \phi) \ (\phi^{-1}\ f_{T_1} \ f_{T_2} \ \phi)$

Surjectivity: Let $f_S$ be an arbitrary element of $SYM(S)$, then we have $f_T = \phi f_S \phi^{-1} \in SYM(T)$ That maps to $f_S$.

Injectivity: let $\phi^{-1}f_{T_1}\phi = \phi^{-1}f_{T_2}\phi$ Then due to the properties of bijections, $f_{T_1} = f_{T_2}$ . $\square$

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Theorem (Cayley’s Theorem): Let $G$ be a group, then $G$ is isomorphic to a subgroup of symmetric group on the set $G$, $Sym(G)$.

Proof: Because left coset maps of a group are bijective, they are a subset of $Sym(G)$. Because left coset maps on a group are isomorphic to the group itself, $G$ is isomorphic to a subgroup of $Sym(G)$. $\square$

https://www.youtube.com/playlist?list=PLA7B08F1D8252DE29 I am planning on writing a sequel with really key ideas of group theory that I have missed here.