Subgroups: Normal and...Abnormal...

1. Definition. Let ${G}$ be a group; denote its group structure as the tuple ${(G,e_{G},\star_{G},\iota_{G})}$ where ${\iota_{G}}$ is the inverse operator, ${e_{G}}$ is the identity element, ${\star_{G}}$ is the binary composition operator.

We define a Subgroup of ${G}$ to consist of a subset ${H\subseteq G}$ equipped with ${G}$'s group structure, i.e.,

1. the binary operation restricted to ${H}$, ${\star|_{H}\colon H\times H\to G}$
2. the inverse operator ${\iota|_{H}\colon H\to H}$
3. the identity element from ${G}$, ${e_{G}\in H}$

such that

1. the group composition is closed on ${H}$, i.e., ${\{h_{1}\star|_{H}h_{2}|h_{1},h_{2}\in H\}\subseteq H}$
2. group inversion is closed in ${H}$, i.e., ${\iota|_{H}(H)\subseteq H}$
3. the usual group properties holds for this group structure induced on ${H}$.

If further ${H\neq G}$, we call ${H}$ a Proper Subgroup of ${G}$.

1.1. Remark. We denote ${H<G}$ if ${H}$ is a proper subgroup of ${G}$, and ${H\leq G}$ if ${H}$ is a generic subgroup of ${G}$.

2. Example (Trivial subgroups). For any group ${G}$, there are always two subgroups available: ${G}$ itself, and the trivial subgroup ${\mathbf{1}}$ consisting of only the identity element. Since these subgroups come "for free", we call them Trivial Subgroups.

3. Example (Subgroups of dihedral group). Let ${n\in\mathbb{N}}$ be greater than 2. Recall the dihedral group ${D_{n}}$ consists of rotations by ${2\pi/n}$ radians and reflections. We have two subgroups (at least): one generated by rotations alone, the other generated by reflections alone.

The subgroup of rotations is a finite group consisting of ${n}$ elements. It's a cyclic group, isomorphic to ${\mathbb{Z}/n\mathbb{Z}}$.

The subgroup of reflections is less exciting. There are two elements in it: reflection about the ${x}$-axis, and the identity transformation. This is a cyclic subgroup of order 2, isomorphic to ${\mathbb{Z}/2\mathbb{Z}}$.

4. Example (NON-Example: subset alone insufficient). Consider modular arithmetic ${\mathbb{Z}/p\mathbb{Z}}$ for some prime ${p}$. This is a group consisting of a set ${\{0,1,\dots,p-1\}}$ equipped with addition modulo ${p}$, inversion maps ${m}$ to ${-m\equiv p-m\bmod{p}}$.

Although the underlying set is a subset of the additive group ${\mathbb{Z}}$, the addition operation is a different function. Consequently, the group structure on ${\mathbb{Z}/p\mathbb{Z}}$ is not equal to the group structure on ${\mathbb{Z}}$. The moral: being a subset alone is insufficient.

5. Example (Symmetric group). Let ${n\in\mathbb{N}}$. The symmetric group ${S_{n}}$ consists of permutations of the set ${\Omega_{n}=\{1,2,\dots,n\}}$. We can consider ${k\leq n}$, and then we have ${S_{k}\leq S_{n}}$ by restricting to permutations of the subset ${\Omega_{k}\subseteq\Omega_{n}}$.

6. Example. Recall the general linear group ${\mathrm{GL}(n,\mathbb{R})}$ is the group of ${n\times n}$ invertible matrices over the real numbers, using matrix multiplication as its binary operator, and matrix inversion for its group inverse. Also recall the determinant $$ \det\colon\mathrm{GL}(n,\mathbb{R})\to\mathbb{R}^{\times} \tag{1}$$ is a group morphism. Its kernel is a subgroup, called the Special Linear Group, denoted $$ \mathrm{SL}(n,\mathbb{R})=\{M\in\mathrm{GL}(n,\mathbb{R})|\det(M)=1\}. \tag{2}$$ The reader can check it is closed under matrix multiplication and matrix inversion.

7. Example. Let ${G}$ be any group. If ${g_{1},g_{2}\in G}$, then their commutator is the element ${[g_{1},g_{2}]=g_{1}g_{2}g_{1}^{-1}g_{2}^{-1}\in G}$. We can construct the Commutator Subgroup of ${G}$, denoted ${[G,G]}$ or (more commonly but more confusingly) by ${G'}$ in the literature, by considering the subgroup generated by commutators of elements of ${G}$. Then an arbitrary element of the commutator subgroup looks like $$ {} [g_{1},g_{2}][g_{3},g_{4}](\dots)[g_{2n-1},g_{2n}]\in[G,G] \tag{3}$$ and we use multiplication from ${G}$.

1. Exercises

Exercise 1. Let ${H_{1}}$, ${H_{2}\leq G}$. Is ${H_{1}\cap H_{2}}$ a subgroup of ${G}$?

Exercise 2. Let ${H_{1}}$, ${H_{2}\leq G}$. Is ${H_{1}\cup H_{2}}$ a subgroup of ${G}$?

Exercise 3. Let ${\varphi\colon G\to K}$ be a group morphism. Is the image of ${\varphi}$ a subgroup of ${K}$?

If ${H<K}$, is its preimage under ${\varphi}$ a subgroup of ${G}$?

Exercise 4. Prove or find a counter-example: if ${G}$ is a cyclic group, then every subgroup is cyclic.

2. Normal Subgroups

8. The special linear group is a rather special situation (no pun intended). We can prove that, for any ${M\in\mathrm{SL}(n,\mathbb{R})}$ and for any ${T\in\mathrm{GL}(n,\mathbb{R})}$, we have ${XMX^{-1}\in\mathrm{SL}(n,\mathbb{R})}$. Really? Look, take its determinant: $$ \begin{split} \det(XMX^{-1})&=\det(X)\det(M)\det(X)^{-1}\\ &=\det(M)=1. \end{split} \tag{4}$$ Its a property shared by the kernel of any group morphism ${\varphi\colon G\to H}$, we'd have ${k\in\ker(\varphi)}$ and ${g\in G}$ satisfy ${\varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(k)=e_{H}}$. This gives us a particular kind of subgroups.

9. Definition. Let ${G}$ be a group. A Normal Subgroup of ${G}$ consists of a subgroup ${H\leq G}$ such that

1. Closed under conjugation by group elements: for any ${h\in H}$ and for any ${g\in G}$, we have ${ghg^{-1}\in H}$.

9.1. Remark. Care must be taken for the normality property: all we ask is for ${ghg^{-1}\in H}$, not that ${ghg^{-1}=h}$.

9.2. Remark (Notation). We denote a normal subgroup ${N}$ of ${G}$ by ${N \triangleleft G}$, and ${N \trianglelefteq G}$ if ${N}$ is a proper normal subgroup of ${G}$. There are a couple ways of remembering which way the triangle points: one is that we just turn the subgroup notation ${N\leq G}$ into a triangle, the other is that it "snitches" on (i.e., points to) the normal subgroup.

10. Example. The trivial subgroups are trivially normal.

11. Example. For the Dihedral group ${D_{n}}$, we have a couple subgroups (one generated by rotations, the other generated by reflections). Is one of them normal?

The rotation subgroup is generated by ${r}$ and satisfies ${r^{n}=1}$. The reflection subgroup is generated by ${s}$ and satisfies ${s^{2}=1}$. Together, they have ${s\circ r^{k}\circ s=r^{-k}}$. This implies the rotation subgroup is normal.

Is the reflection subgroup normal?

12. Theorem. Let ${G}$ be a group. If ${G}$ is Abelian, then every subgroup ${H\leq G}$ is normal.

Proof: Assume ${G}$ is Abelian. Let ${H\leq G}$ be an arbitrary subgroup of ${G}$, let ${g\in G}$ and ${h\in H}$ be arbitrary group elements. Then conjugation looks like ${g+h-g=h}$. Hence ${H}$ is a normal subgroup of ${G}$ by Definition 9. ∎

13. Example (Converse to Theorem 12 is false). Consider the quaternion group ${Q_{8}}$ generated by ${i^{4}=j^{4}=k^{4}=(ijk)^{2}=1}$ and so ${ij=k}$, ${jk=i}$, ${j=ki}$. We really have only 8 elements in ${Q_{8}}$: ${1}$, ${-1}$, ${i}$, ${-i}$, ${j}$, ${-j}$, ${k}$, ${-k}$. But we only need two of the "purely imaginary" elements (${i}$, ${j}$, ${k}$) to generate the entire group.

The proper subgroups would be generated from one complex generator, or from ${-1}$. We claim they are all normal. The case of ${\{\pm1\}<Q_{8}}$ is obviously normal, since ${x(-1)x^{-1}=-1}$ for any ${x\in Q_{8}}$.

The reasoning for proper subgroups generated by one imaginary element resembles one another, so let's consider the subgroup generated by ${i}$. Then we see $$ jij^{-1}=-jk=-i \tag{5}$$ and $$ j^{3}ij^{-3}=(-j)i(j)=i. \tag{6}$$ Similarly we'd find, from ${i^{3}=-i}$, that $$ ji^{3}j^{-1}=i \tag{7}$$ and $$ j^{3}i^{3}j^{-3}=-i. \tag{8}$$ Thus the subgroup generated by ${i}$ is closed under conjugation from all elements from ${Q_{8}}$.

Really? Well, any generic element in ${Q_{8}}$ looks like ${i^{m}j^{n}}$. Its inverse would be ${(i^{m}j^{n})^{-1}=j^{-n}i^{-m}}$. So conjugation by these elements would amount to multiplying the element ${i^{\ell}}$ from the subgroup by some ${i^{q}}$. But that's okay: elements of the form ${i^{q}}$ belong to the subgroup anyways.

14. Theorem. If ${\varphi\colon G\to H}$ is a group morphism, then ${\ker(\varphi)}$ is a normal subgroup of ${G}$.

The proof was sketched earlier in §8.

15. Theorem. Let ${G}$ be a group. Every normal subgroup ${N\triangleleft G}$ is the kernel of some group morphism.

15.1. Remark. We do not yet have the technology to prove this, yet, but it is true. Basically, we construct a morphism by taking ${g\in G}$ and mapping it to cosets ${gN=\{gn\in G|n\in N\}}$. The difficulty we have is that the image of this mapping is the collection of left cosets of ${N}$, which may or may not be a group. Further, it was rather arbitrary mapping it to left cosets of ${N}$: why not map ${g}$ to ${Ng=\{ng\in G|n\in N\}}$? Wouldn't these be distinct (i.e., not equal) mappings?

3. Exercises

Exercise 5. Find the normal subgroups of the symmetric group ${S_{n}}$.

Exercise 6. Let ${G}$ be a group, ${H\triangleleft G}$ be a normal subgroup, and ${N\triangleleft H}$ be a normal subgroup of ${H}$. Is ${N}$ a normal subgroup of ${G}$?

Exercise 7. Let ${G}$ be a group, recall the commutator subgroup from Example 7 that ${[G,G]}$ is a subgroup of ${G}$. Prove or find a counter-example: the commutator subgroup is a normal subgroup of ${G}$.

Exercise 8. Let ${\varphi\colon G\to H}$ be a group morphism, ${N\triangleleft G}$ be a normal subgroup. Is ${\varphi(N)}$ a normal subgroup of ${G}$? Consider the cases when ${\varphi}$ is injective, and when ${\varphi}$ is surjective.

Exercise 9. Let ${\varphi\colon G\to H}$ be a group morphism, and ${N_{H}\triangleleft G}$ be a normal subgroup. Is the preimage ${\varphi^{-1}(N_{H}) = \{g\in G|\varphi(g)\in N_{H}\}}$ a normal subgroup of ${G}$? Is it even a subgroup?

Group Morphisms

1. There is some "Kabuki theater" whenever introducing a new mathematical gadget, thanks to category theory: we have our new gadget, then we could ask about morphisms (very important), "subgadgets", and universal constructions (products, quotients, etc.). The exciting thing, as in Kabuki theater, is the order and manner of presentation.

2. Definition. Let ${G}$ and ${H}$ be groups. We define a Group Morphism to consist of a function ${\varphi\colon G\to H}$ of the underlying sets such that

1. group operation is preserved: for any ${g_{1}}$, ${g_{2}\in G}$, we have ${\varphi(g_{1}g_{2})=\varphi(g_{1})\varphi(g_{2})}$;
2. identity element is preserved: if ${e_{G}\in G}$ is the identity element of ${G}$ and ${e_{H}\in H}$ is the identity element of ${H}$, then ${\varphi(e_{G})=e_{H}}$;
3. inverse is preserved: for any ${g\in G}$, ${\varphi(g^{-1})=\varphi(g)^{-1}}$.

2.1. Remark. Older texts refer to group morphisms as "group homomorphisms". After category theory became popular and part of the standard curriculum, the "homo-" prefix was dropped because group morphisms live in the same category, so it was redundant.

3. Example. One "low hanging fruit" for morphism examples is the identity morphism. We should check, for any group ${G}$, the identity function ${\mathrm{id}\colon G\to G}$ is a bona fide group morphism.

We see it preserves the group operation. For any ${g_{1}}$, ${g_{2}\in G}$, we have ${\mathrm{id}(g_{1}g_{2})=g_{1}g_{2}}$ by definition of the identity function. But this is also equal to ${\mathrm{id}(g_{1})\mathrm{id}(g_{2})}$. Thus the group operation is preserved.

The identity element is preserved ${\mathrm{id}(e_{G})=e_{G}}$.

Inversion is also preserved ${\mathrm{id}(g^{-1})=g^{-1}=\mathrm{id}(g)^{-1}}$ for any ${g\in G}$.

Thus taken together, it follows the identity mapping satisfies the axioms of a group morphism.

4. Example. Let ${G}$ be any group, and consider ${\mathbb{Z}}$ equipped with addition as a group. For each ${g\in G}$, we have a group morphism ${\varphi\colon\mathbb{Z}\to G}$ sending ${1\in\mathbb{Z}}$ to ${g\in G}$. Is this really a group morphism?

We can check that the properties are (or, ought to be) satisfied. If the group operation is preserved, then ${\varphi(1+1)=\varphi(1)\varphi(1)=g^{2}}$ and more generally, for any ${m\in\mathbb{Z}}$, we have ${\varphi(m+1)=\varphi(1)^{m}=g^{m}}$.

For the identity element being preserved, that means ${\varphi(0)=e_{G}}$, which is fine: it corresponds to ${g^{0}=e_{G}}$.

Group inverses would be ${\varphi(-m)=\varphi(m)^{-1}=(g^{m})^{-1}}$. And we know this is precisely the same as ${g^{-m}}$.

5. Example. Consider the group ${\mathrm{GL}(2,\mathbb{R})}$ and the multiplicative group ${\mathbb{R}^{\times}}$ of nonzero real numbers. Then the determinant $$ \det\colon\mathrm{GL}(2,\mathbb{R})\to\mathbb{R}^{\times} \tag{1}$$ is a group morphism. Let us prove it!

We see, for any matrices ${M}$, ${N\in\mathrm{GL}(2,\mathbb{R})}$ we have $$ \det(MN)=\det(M)\det(N). \tag{2}$$ This is a familiar fact in linear algebra. But for us, it tells us the group operation is preserved.

The identity element must be mapped to the identity element. We see the identity matrix ${I\in\mathrm{GL}(2,\mathbb{R})}$ has ${\det(I)=1}$. Thus the determinant preserves the group identity element.

As far as the group inverse, well, this follows from previous results, right? After all, if ${M\in\mathrm{GL}(2,\mathbb{R})}$, then ${M^{-1}\in\mathrm{GL}(2,\mathbb{R})}$, and $$ I = M^{-1}M \tag{3}$$ so $$ \det(M^{-1}M)=\det(M^{-1})\det(M)=1 \tag{4}$$ and thus by division $$ \det(M^{-1})=\det(M)^{-1}. \tag{5}$$ Thus the group inverse operator is preserved.

6. Definition. Let ${\varphi\colon G\to H}$ be a group morphism. We define the Kernel of ${\varphi}$ to be the pre-image of the identity element of ${H}$: \begin{equation*} \ker(\varphi)=\{g\in G|\varphi(g)=e_{H}\}. \end{equation*}

7. Example. For the group morphism ${\det\colon\mathrm{GL}(2,\mathbb{R})\to\mathbb{R}^{\times}}$, the kernel would be $$ \ker(\det)=\{M\in\mathrm{GL}(2,\mathbb{R})|\det(M)=1\}. \tag{6}$$ That is to say, it consists of matrices with unit determinant. Observe, this is a group under matrix multiplication: if two matrices have unit determinant, their product has unit determinant; the identity matrix is in the kernel; and it's closed under inverses. This is an important group called the Special Linear Group, denoted ${\mathrm{SL}(2,\mathbb{R})}$.

1. Properties of Morphisms

8. Proposition. Let ${\varphi\colon G\to H}$ be a group morphism. If ${g\in G}$ is any element, then ${\varphi(g^{-1})=\varphi(g)^{-1}}$.

Proof: Let ${g\in G}$ (so ${\varphi(g)\in H}$). We find ${\varphi(g\cdot g^{-1})=\varphi(g)\varphi(g^{-1})=e_{H}}$, thus multiplying on the left by ${\varphi(g)^{-1}}$ gives the result. ∎

9. Proposition. Let ${\varphi\colon G\to H}$ be a group morphism. If ${g\in G}$ is any element and ${n\in\mathbb{Z}}$ is any integer, then ${\varphi(g^{n})=\varphi(g)^{n}}$.

Proof: Per cases since ${n<0}$ or ${n=0}$ or ${n>0}$. The ${n=0}$ case is obvious.

For ${n>0}$, by induction. The base case ${n=1}$ gives ${\varphi(g^{1})=\varphi(g)^{1}}$, which is obvious. Assume this holds for arbitrary ${n}$. Then the inductive case ${n+1}$ is $$ \varphi(g^{n+1})=\varphi(g^{n}g)=\varphi(g)^{n}\varphi(g)=\varphi(g)^{n+1}. \tag{7}$$ Thus we have proven the result for non-negative ${n}$.

For negative ${n\in\mathbb{Z}}$, the proof is analogous. ∎

10. Theorem. The composition of group morphisms is a group morphism. More explicitly, if ${\varphi\colon G\to H}$ and ${\psi\colon H\to K}$ are group morphisms, then ${\psi\circ\varphi\colon G\to K}$ is a group morphism.

11. Theorem. Let ${\varphi\colon G\to H}$ be a group morphism. If ${\ker(\varphi)=\{e_{G}\}}$, then ${\varphi}$ is injective.

Proof: Assume ${\ker(\varphi)=\{e_{G}\}}$. Let ${g_{1}}$, ${g_{2}\in G}$ be completely arbitrary. (We want to show if ${\varphi(g_{1})=\varphi(g_{2})}$, then ${g_{1}=g_{2}}$.) Assume ${\varphi(g_{1})=\varphi(g_{2})}$. Then ${\varphi(g_{1})\varphi(g_{2})^{-1}=e_{H}}$ by multiplying both sides on the right by ${\varphi(g_{2})^{-1}}$. And ${\varphi(g_{1})\varphi(g_{2}^{-1})=\varphi(g_{1}\cdot g_{2}^{-1})=e_{H}}$. Thus ${g_{1}\cdot g_{2}^{-1}\in\ker(\varphi)}$ by definition of the kernel. But we assumed the only member of the kernel was identity element. Thus ${g_{1}\cdot g_{2}^{-1}=e_{G}}$, and moreover ${g_{1}=g_{2}}$. Hence ${\varphi}$ is injective. ∎

2. Exercises

Exercise 1. Let ${G}$ be a group, ${n\in\mathbb{N}}$ be a fixed positive integer. Prove or find a counter-example: ${\varphi\colon G\to G}$, sending ${g}$ to ${\varphi(g)=g^{n}}$ is a group morphism.

Exercise 2. Prove or find a counter-example: the matrix trace ${\mathop{\rm tr}\nolimits\colon\mathrm{GL}(2,\mathbb{R})\to\mathbb{R}}$ is a group morphism.

Exercise 3. Is the exponential function on the real numbers a group morphism ${\exp\colon\mathbb{R}\to\mathbb{R}^{\times}}$?

Exercise 4. Is the matrix exponential a group morphism ${\exp\colon\mathrm{Mat}_{2}(\mathbb{R})\to\mathrm{GL}(2,\mathbb{R})}$? [Hint: $\mathrm{Mat}_{2}(\mathbb{R})$ has its group operation be matrix addition. Is this preserved?]

Introducing Groups to Beginners

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1. Introduction. We will do some group theory. Here "group" refers to a "group of symmetry transformations", and we should think of elements of the group as functions mapping an object to itself in some particularly symmetric way.

2. Definition. A Group consists of a set ${G}$ equipped with

1. a law of composition ${\circ\colon G\times G\to G}$,
2. an identity element ${e\in G}$, and
3. an inverse operator ${(-)^{-1}\colon G\to G}$

such that

1. Associativity: For any ${g_{1}}$, ${g_{2}}$, ${g_{3}\in G}$, ${(g_{1}\circ g_{2})\circ g_{3}=g_{1}\circ(g_{2}\circ g_{3})}$
2. Unit law: For any ${g\in G}$, ${g\circ e=e\circ g=g}$
3. Inverse law: For any ${g\in G}$, ${g^{-1}\circ g=g\circ g^{-1}=e}$.

3. Effective Thinking Principle: Create Examples. Whenever encountering a new definition, it's useful to construct examples. Plus, it's fun. Now let us consider a bunch of examples!

4. Example (Trivial). One strategy is to find the most boring example possible. We can't use ${G=\emptyset}$ since a group must contain at least one element: the identity element ${e\in G}$. Thus the next most boring candidate is the group containing only the identity element ${G=\{e\}}$. This is the Trivial Group.

5. Example (Dihedral). Consider the regular ${n}$-gon in the plane ${X\subset\mathbb{R}^{2}}$ with vertices located at ${(\cos(k2\pi/n), \sin(k2\pi/n))}$ for ${k=0,1,\dots,n-1}$. We also require ${n\geq3}$ to form a non-degenerate polygon (${n=2}$ is just a line segment, and ${n=1}$ is one dot).

We can rotate the polygon by multiples of ${2\pi/n}$ radians. There are several ways to visualize this, I suppose we could consider rotations of the plane by ${2\pi/n}$ radians: $$ r\colon\mathbb{R}^{2}\to\mathbb{R}^{2} \tag{1}$$ which acts like the linear transformation $$ r \begin{pmatrix} x\\ y \end{pmatrix} := \begin{pmatrix} \cos(2\pi/n) & -\sin(2\pi/n)\\ \sin(2\pi/n) & \cos(2\pi/n) \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}. \tag{2}$$ We see that the image of our ${n}$-gon under this transformation ${r(X)=X}$ remains invariant.

The other transformation worth exploring is reflecting about the ${x}$-axis, ${s\colon\mathbb{R}^{2}\to\mathbb{R}^{2}}$ which may be defined by $$ s \begin{pmatrix} x\\ y \end{pmatrix} := \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}. \tag{3}$$ This transformation also leaves our polygon invariant ${s(X)=X}$.

We can compose these two types of transformations. Observe that ${s\circ s=\mathrm{id}}$ and the ${n}$-fold composition ${r^{n}=r\circ\dots\circ r=\mathrm{id}}$ both yield the identity transformation ${\mathrm{id}(x)=x}$ for all ${x\in\mathbb{R}^{2}}$. Then we have ${2n}$ symmetry transformations: ${\mathrm{id}}$, ${r}$, ..., ${r^{n-1}}$; and ${s}$, ${s\circ r}$, ..., ${s\circ r^{n-1}}$. What about, say, ${r\circ s}$? We find ${s\circ r^{k}\circ s=r^{-k}}$, so ${r^{k}\circ s = s\circ r^{-k}}$. Thus it's contained in our list of symmetry transformations.

The symmetry group thus constructed is called the Dihedral Group. Geometers denote it by ${D_{n}}$, algebraists denote it by ${D_{2n}}$, and we denote it by ${D_{n}}$.

6. Example (Rotations of regular polygon). We can restrict our attention, working with the previous example further, to only rotations of the regular ${n}$-gon by multiples of ${2\pi/n}$ radians. We can describe this group as "generated by a single element", i.e., symmetries are of the form ${r^{k}}$ for ${k\in\mathbb{Z}}$. This is an example of a Cyclic Group. In particular, it is commutative: any symmetries ${r_{1}}$ and ${r_{2}}$ satisfy ${r_{1}\circ r_{2}=r_{2}\circ r_{1}}$. These are special situations, let us carve out space to define these concepts explicitly.

7. Definition. We call a group ${G}$ Abelian if it is commutative, i.e., for any transformations ${f}$, ${g\in G}$ we have ${f\circ g = g\circ f}$. In this case, we write ${f\circ g}$ as ${f+g}$, using the plus sign to stress commutativity.

8. Definition. We call a group ${G}$ Cyclic if there is at least one element ${g\in G}$ such that ${\{g^{n}\mid n\in\mathbb{Z}\}=G}$ the entire group consists of iterates of ${g}$ and ${g^{-1}}$.

9. Example (Number Systems). Another few examples the reader may know are the familiar number systems under addition: the integers ${\mathbb{Z}}$, the rational numbers ${\mathbb{Q}}$, the real numbers ${\mathbb{R}}$, and the complex numbers ${\mathbb{C}}$. They are commutative groups.

10. Example (Infinite dihedral). We can take the infinite limit of the dihedral group to get the infinite dihedral group ${D_{\infty}}$. We formally describe it as consisting of "rotations" ${r}$ and "reflections" ${s}$ such that

1. ${r^{m}\circ r^{n} = r^{m+n}}$ for any ${m}$, ${n\in\mathbb{Z}}$;
2. ${s\circ r^{m}\circ s = r^{-m}}$ for any ${m\in\mathbb{Z}}$;
3. ${s\circ s = e}$;
4. ${r^{n}\circ r^{-n} = r^{-n}\circ r^{n} = e}$ for any ${n\in\mathbb{Z}}$, in particular ${r^{0}=e}$.

In this sense, the "infinite limit" turns rotations into something like the integers.

11. Example (Circular dihedral). A more intuitive "infinite limit" of the dihedral group is the symmetries of the unit circle ${S^{1}}$ in the plane ${\mathbb{R}^{2}}$. These are anti-clockwise rotations and reflection about the ${x}$-axis, but rotations are parametrized by a real parameter (the "angle"): $$ r_{\theta}~``=\!\!\mbox{"} \begin{pmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}. \tag{4}$$ Here we write an "equals" sign in quotes because this is the intuition. A group is abstract, whereas the matrix is a concrete realization of the symmetry.

The reader should verify the axioms for a group are satisfied, with the hint that ${r_{\theta}\circ r_{\phi} = r_{\theta+\phi}}$ and the usual relation between reflection and rotation holds.

This group is called the Orthogonal Group in 2-dimensions.

Exercises

Exercise 1. Is ${\mathbb{Z}}$ a cyclic group? Is ${\mathbb{C}}$ a cyclic group?

Exercise 2. Is the non-negative integers ${\mathbb{N}_{0}}$ a group under addition? Under multiplication?

Exercise 3. Are the positive real numbers ${\mathbb{R}_{\text{pos}}}$ a group under multiplication?

Exercise 4. Pick your favorite polyhedron in 3-dimensions. Determine its symmetry group.

Exercise 5. Complex conjugation acts on ${\mathbb{C}}$ by sending ${x+i\cdot y}$ to ${x-i\cdot y}$. Does this give us a symmetry group?

Exercise 6 (challenging). If we consider polynomials with coefficients in, say, rational numbers (denoted ${\mathbb{Q}[x]}$ for polynomials with the unknown ${x}$), then how can we form a symmetry group of ${\mathbb{Q}[x]}$?

Exercise 7 (General Linear Group). Take ${n\in\mathbb{N}}$ to be a fixed positive integer, preferably ${n\geq2}$. Consider the collection of invertible ${n}$-by-${n}$ matrices with entries which are rational numbers $${\mathrm{GL}(n, \mathbb{Q}) = \{ M\in\mathrm{Mat}(n\times n, \mathbb{Q}) \mid \det(M)\neq0\}.}$$ Prove this is a group under matrix multiplication.