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.,
- the binary operation restricted to ${H}$, ${\star|_{H}\colon H\times H\to G}$
- the inverse operator ${\iota|_{H}\colon H\to H}$
- the identity element from ${G}$, ${e_{G}\in H}$
- the group composition is closed on ${H}$, i.e., ${\{h_{1}\star|_{H}h_{2}|h_{1},h_{2}\in H\}\subseteq H}$
- group inversion is closed in ${H}$, i.e., ${\iota|_{H}(H)\subseteq H}$
- the usual group properties holds for this group structure induced on ${H}$.
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
- 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?
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