Group Theory, Part I: Definitions and Basics

September 19th, 2017

Introduction

A group is a set together with a binary operation (multiplication) so that:

The order of an element $a$ is the minimum integer $n$ so that $a^n = e$. The subgroup consisting of all elements of the group of finite order is called the torsion subgroup.

Example

An important example of a group is the dihedral group $D_n$. It is generated by two kinds of elements: rotations, and reflections. It describes the symmetries of an $n$-gon with composition. The two kinds of elements are respectively described as:

$$ r^n = e \\ s^2 = e \\ srs = r^{-1} $$

$D_1$ is for example defined as $\{1,r\}$ so it is simply $\mathbb{Z}/2\mathbb{Z}$. On the other hand, $D_2 = \{1,r,s,rs\}$ is not cyclic; it is called the Klein group or the 4-group, which is distinct from $\mathbb{Z}/4\mathbb{Z}$.

The General Linear Group

An important group is the general group $GL(V)$. For an $n$-dimensional vector space $V$ over a field, we can think of $GL(V)$ as the set of $n\times n$ matrices over a field with nonzero determinant – with multiplication defined in the usual way (once we fix a basis).

A bilinear form $\phi: V\times V \rightarrow F$ that is linear in each variable. An automorphism of $\phi$ is is an isomorphism $\alpha: V \rightarrow V$ so that:

$$ \phi(\alpha v, \alpha w) = \phi(v,w) $$

With a choice of a basis, we can restate this condition in terms of the matrix for $\alpha$ and the matrix $P$ for $\phi$:

$$\begin{aligned} (Av)^T\cdot P Aw &= v^T P w \\ v^TA^TPAw &= v^T P w \end{aligned}$$

So:

$$ A^TPA = P $$

In particular, if $\phi$ is symmetric, i.e.:

$$ \phi(v,w) = \phi(w,v) $$

Then we have the following definition.

Definition: For a symmetric non-degenerate bilinear form $\phi$, define its automorphism group $Aut(\phi)$ to be the isomorphisms $\alpha$ so that $\phi(\alpha v, \alpha w) = \phi(v,w)$. This is called the orthogonal group of $\phi$.

Definition: For a skew-symmetric non-degenerate bilinear form $\phi$, define its automorphism group $Aut(\phi)$ to be the isomorphisms $\alpha$ so that $\phi(\alpha v, \alpha w) = \phi(v,w)$. This is called the symplectic group of $\phi$.

In this case, we can write $\phi$ in some basis as the matrix:

$$J_{2m}= \begin{bmatrix} 0 & I_m \\ -I_m & 0 \end{bmatrix} $$

Where $2m=n$. Therefore, the symplectic group condition simply means a matrix has the property:

$$ A^TJ_{2m}A = J_{2m} $$

Subgroups

A subgroup is a subset of a group which is closed under multiplication and inverses, and which contains the identity. A particularly important is called the center of a group.

Definition: The center of a group $G$, denoted $Z(G)$ consists of all the elements which commute with all of $G$, i.e.:

$$ Z(G) = \{z \in G\ :\ zx=xz\ \forall x\in G\} $$

Proposition

An intersection of subgroups is a subgroup.

The proof here is fairly straightforward.

We can talk about the cosets of a subgroup $H$ as elements of the form $aH$ for some $a\in G$, where:

$$ aH = \{ah\ :\ h \in H\} $$

Cosets are well-defined, and are either disjoint or equal. Suppose that $a\ in bH$, then we can say for some $h\in H$:

$$ a = bh \\ aH = bhH = bH $$

So that means we can write a coset as $aH$ for any choice of representative $a$. By the above argument, if two cosets share a single element, they are the same set. Finally, we can map $aH$ to $bH$ via multiplication by $ba^{-1}$ (and conversely, map from $bH$ to $aH$ via multiplication by $ab^{-1}$). Thus, all the cosets are the same size.

Definition: The index of a subgroup $H$ of $G$ is the number of left cosets of $H$ in $G$, and is denoted $(G:H)$.

Proposition (Lagrange’s Theorem)

The order of a subgroup divides the order of the group.

We have:

$$ |G| = (G:H)|H| $$

Therefore, $|H|$ divides $|G|$.

As a corollary, we consider the group generated by a certain element $a$. It has size $n$, where $n$ is the order of $a$, and forms a subgroup. Thus, the order of any element in a group divides the order of the group.

We also have the following “cancellation” theorem. If $H$ is a subgroup of $G$ and $K$ is a subgroup of $H$, we have:

$$ (G:K) = (G:H)(H:K) $$

Homomorphisms

Definition: A homomorphism between groups $G, G’$ is a map $\varphi: G\rightarrow G’$ so that $\varphi(ab) = \varphi(a)\varphi(b)$. In a sense, a homomorphism preserves the structure of the group. If a homomorphism is bijective, we say that it is an isomorphism.

Cayley’s Theorem

An important theorem is Cayley’s Theorem, which says we can think of each group as a subgroup of a permutation group. For $a \in G$, define the map:

$$ \begin{aligned} \phi_a: G &\rightarrow G \\ \phi_a(b) &= ab \end{aligned}$$

Thus, the map $\phi_a$ is just multiplication by $A$. We can also show that it is a bijection, since we have:

$$ \phi_a\circ\phi_{a^{-1}}(b) = \phi_a(a^{-1}b) = aa^{-1}b = b $$

And in fact we can say that:

So this brings us to Cayley’s Theorem:

Any finite group is a subgroup of a symmetric group.

Normal Subgroups

Definition: A subgroup $N$ of a group $G$ is normal if $gNg^{-1} = N$ for all $g\in G$. A normal subgroup is denoted $N \trianglelefteq G$.

It is sufficient to check that $gNg^{-1} \subset N$ for each $g$, since multiplying gives us $Ng^{-1} = g^{-1}N \implies N \subseteq g^{-1}Ng$, and substituting $g=g^{-1}$ we get the reverse inclusion.

Note however, that we can find a subgroup $N$ and an element $g$ so that $gNg^{-1} \subset N$ with strict inequality; however, if this holds for all $g$, then we indeed have a normal subgroup.

Proposition

Every subgroup of index two is normal.

Suppose $H$ is a subgroup of index two. Pick $g\in G$ which is not in $H$. then $gH$ is the complement of $H$. Similarly, $Hg$ is the complement of $H$. So we have $gH = Hg$. Then $gHg^{-1} = H$.

Definition: A group is simple if it has no normal subgroups other than itself and the trivial subgroup.

Proposition

Suppose $H$, $N$ are subgroups of $G$ and $N$ is a normal subgroup. Then $HN = {hn\ :\ h\in H, n \in N}$ is a subgroup of $G$. If $H$ is also a normal subgroup, then $HN$ is a normal subgroup of $G$.

Note that $gNg^{-1} = N$, so that we can write $gN = Ng$. For any $n\in N$, we can write $gn = n’g$ where $n’ \in N$.

Taking $h_1n_1, h_2n_2 \in HN$, we have:

$$ (h_1n_1)(h_2n_2) = h_1h_2n_1'n_2 \in HN $$

So indeed $HN$ is closed under multiplication. It contains the identity automatically, and we can check inverses:

$$ (hn)^{-1} = n^{-1}h^{-1} = h^{-1}n'^{-1} \in HN $$

So indeed $HN$ is a subgroup.

If $H,N$ are both normal, we can write:

$$ gHNg^{-1} = gHg^{-1}gNg^{-1} = HN $$

And we are done. We can also define the normal subgroup generated by any set in $G$.

Definition: For any set $X \subset G$, the smallest normal subgroup generated by $X$ is exactly:

$$ \bigcup\limits_{g\in G} gXg^{-1} $$

Theorem

A subgroup $N$ of $G$ is normal iff it is the kernel of some homomorphism.

Evidently, the kernel of a homomorphism is a normal subgroup since for any $x\in \ker \varphi$:

$$ \varphi(gxg^{-1})=\varphi(g)e\varphi(g)^{-1} = e $$

Conversely, we map $g \mapsto gN$, i.e. map to cosets. We just need to show that $G/N$ has a group structure which is preserved by this map. Define $(aN)(bN) = (ab)N$. We need to show that this multiplication is well defined.

Suppose that $aN = a’N$ and $bN = b’N$. Then we can show:

$$ abN = a(bN) = ab'N = aNb' = a'Nb' =a'b'N $$

Where we use freely here that $aN=Na$ by the fact that $N$ is a normal subgroup. So indeed this map is well defined, and preserves the group structure, and its kernel is evidently $N$. We call $G/N$ the quotient of $G$ by $N$.

The Isomorphism Theorems

As per usual, we have the isomorphism theorems.

First Isomorphism Theorem

Let $\varphi: G \rightarrow G’$ be a homomorphism of groups. Then:

$$ \frac{G}{\ker \varphi} \cong \varphi(G) $$

And since $\ker \varphi$ is a normal subgroup by the above discussion, we have that $\varphi(G)$ is a subgroup of $G’$.

Second Isomorphism Theorem

Let $S$ be a subgroup of $G$, and $N$ a normal subgroup of $G$. Then:

Third Isomorphism Theorem

Suppose $K,N$ are normal subgroups of $G$ with $N\subseteq K \subseteq G$. Then:

$$\frac{G/N}{K/N} \cong \frac{G}{K}$$

Furthermore, we have the following correspondences from the third isomorphism theorem:

“Fourth” Isomorphism Theorem

Suppose $N$ is a normal subgroup of $G$. Then there is a correspondence between subgroups $K$ of $G$ which contain $N$ and subgroups of $G/N$, given by:

$$K \leftrightarrow kN$$

Where $k \in K$ is a representative. Similarly, the same bijection gives a correspondence between normal subgroups $K$ of $G$ which contain $N$ and normal subgroups of $G/N$.

Endomorphisms of Abelian Groups

Definition: An endomorphism is a homomorphism from a group to itself.

Definition: An automorphism is an isomorphism from a group to itself, i.e. a bijective endomorphism.

For any group, there is at least one endomorphism: the so-called trivial endomorphism, which sends every element to $0$.

In particular, suppose that we have an abelian group, and two endomorphisms $f,g \in End(G)$. Then we have for fixed $a\in G$:

$$ (f+g)(a+b) = f(a+b) + g(a+b) = f(a) + f(b)+g(a) + g(b) $$

Now, as a consequence of the fact that $G$ is abelian, we then can write:

$$ (f+g)(a+b) = f(a) + g(a) + f(b) + g(b) = (f+g)(a)+(f+g)(b) $$

And so $f+g\in End(G)$. Thus, $End(G)$ is in particular a group (since it is closed under addition). Indeed it is a ring as well, with multiplication as composition. With some work, in some cases we can think of an endomorphism ring as a ring of matrices. In all cases, we can think of rings as subrings of the endomorphism ring of some underlying abelian group (in fact, itself considered as a group).

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