Subgroups

Subgroups are subsets of groups which are also groups in their own right.

Definition and Properties

Definition

Let (G,\star) be a group. A subgroup of (G,\star) is a group (H,\star) such that H \subseteq G.

If (H,\star) is a subgroup of (G,\star), we write

(H,\star) \leq (G,\star)

Example

Let G be the group (\mathbb{Q},+). We previously showed that (\mathbb{Z},+) was a group. Because \mathbb{Z} \subseteq \mathbb{Q}, we have

(\mathbb{Z},+) \leq (\mathbb{Q},+)

Theorem

Let (G,\star) be a group, and H \subseteq G. Then (H,\star) is a subgroup of (G,\star) if and only if

(1) For all x,y \in H, x \star y \in H.

(2) The identity element e of (G,\star) is in H.

(3) For all a \in H, a^{-1} \in H.

Trivially, if G is a group, H = G and H = e_G are both subgroups of G.

Example

A slightly harder example, if m \in \mathbb{Z}, then

m\mathbb{Z} := \{ma \mid a \in \mathbb{Z}\} \leq (\mathbb{Z},+).

Lets define

\mathbb{Z}^+ := \{n \in \mathbb{Z} \mid n > 0\}.

It is clear that

(\mathbb{Z}^+,+) \not \leq (\mathbb{Z},+),

As \mathbb{Z}^+ has no identity element, or any inverses.

The relation \leq (is subgroup of) is transitive. The relation is also reflexive, but the relation is not symmetric.

Proposition

If H, K are both subgroups of G, then H \cap K is a subgroup of G.

Proof (FINISH)

We can prove that H \cap K fulfills the three properties of being a subgroup.

(1)\, e \in H, e \in K \implies e \in H \cap K

(2)\,x,y \in H \cup K \implies x \star y \in H \cup

There is a simpler way however to check if a group H is a subgroup of G, called the Subgroup criterion.

Theorem (Subgroup criterion)

A group H is a subgroup of G if

• H \not = \varnothing
• x,y \in H \implies x \star y^{-1} \in H

Proof (Subgroup criterion)

If H is a subgroup of G then e \in H, thus H \not = \varnothing, and if x,y \in H, then y^{-1} is in H. Thus x \star y^{-1} is in H.

Now if H satisfies the two properties, we have some x \in H, as H is not empty. If y = x, then

x,x \in H \implies x \star x^{-1} \in H \implies e \in H

And then with e,x

e,x \in H \implies e \star x^{-1} \in H \implies x^{-1} \in H

If x,y \in H then y^{-1} \in H. Thus

x,y^{-1} \in H \implies x \star (y^{-1})^{-1} \in H \implies x \star y \in H

So H is a group.

Centralizers, Normalizers, Center

Definition

Let A be a subset of a group G.

The Centralizer of A in G is a set of elements of G that commute with all elements of A.

C_G(A) = \{g \in G \mid gag^{-1} = a, \; \forall a \in A\}

Theorem

The centralizer of a subset A of G is a subgroup of G. C_G(A) \leq G

Proof

We know

e \in C_G(A)

Thus we can fulfill the first property of a subgroup C_G(A) \not = \varnothing. Now, let x,y \in C_G(A). We see

y^{-1}yayy^{-1} = y^{-1}ay

a = y^{-1}ay = y^{-1}a(y^{-1})^{-1}

Thus y^{-1} \in C_G(A).

Now we wish to show xy \in C_G(A)

(xy)a(xy)^{-1} = xyay^{-1}x^{-1}

Because x,y \in C_G(A), we have xy \in C_G(A). Thus

C_G(A) \leq G

Attempting to give an explanation, the centralizer is an operation which when given a set G, and a subset A \subset G, gives all elements of A which commute with every element in G.

Definition

The center of G is defined

Z(G) = \{g \in G \mid gx = xg,\, \forall x \in G\}

The center of G is the centralizer where A = G. Clearly Then

Z(G) = C_G(G)

Z(G) \leq G

Notation

Let A be a subset of a group G, such that g \in G.

gAg^{-1} = \{gag^{-1} \mid a \in A\}

Note that gag^{-1} does not necessarily have to equal a. Given two elements a,b \in A, such that gag^{-1} = b\quad gbg^{-1} = a

We would still write gAg^{-1} = A.

Definition

The normalizer of A in G is denoted

N_G(A) = \{g \in G \mid gAg^{-1} = A\}

Note that C_G(A) \leq N_G(A).

If G is abelian, then

Z(G) = G, \quad C_G(A) = G, \quad N_G(A) = G

Example (CHECK)

Let G = D_8, and our subset A = \{e,r,r^2,r^3\}.

We want to find C_{D_8}

We know that A \subset C_{D_8}(A) because rotations are commutative, that meaning

r^ir^j = r^{i+j} = r^{j+i} = r^jr^i

We know that sr \not = rs, so s\not \in C_{D_8}(A). We now need to check the remaining elements of D_8, sr,sr^2,sr^3.

Because C_{D_8}(A) \leq D_8 we have

sr^i \in C_{D_8}(A) \implies sr^ir^{-1} = s \in C_{D_8}

Which is a contradiction, thus

C_{D_8}(A) = A

Now lets find N_{D_8}(A).

Because C_{D_8} \leq N_{D_8}, we know A \subseteq N_{D_8}(A). We have then

sAs^{-1} = \{ses^{-1},srs^{-1},sr^2s^{-1},sr^3s^{-1}\}

Take as a fact r^is = sr^{-1}. Thus

sAs^{-1} = \{e,r^3,r^2,r\}

Thus s \in N_{D_8}. We need now check sr^i. Because N_{D_8}(A) \leq D_8, we have that sr^i is in the normalizer. Thus

N_{D_8}(A) = D_8

We now wish to find the center of D_8. Since

Z(D_8) \leq C_{D_8}(A)

We know that Z(D_8) \subseteq A.

We can check these four elements of A to see if their commutative with everything. We have

rs = sr^{-1} \not =sr

Thus r is not commutative with s, and is not in the center.

r^2s = sr^{-2} = sr^2

So r^2 commutes with s. Now we check if it commutes with sr^i.

r^2sr^i = sr^{-2}r^i = sr^ir^{-2}

Thus our center is Z(D_8) = \{e,r^2\}

Cyclic groups (REWRITE)

Definition

A group H is called cyclic if it is generated by one element. We denote H

H = \langle x \rangle = \{x^n \mid n \in \mathbb{Z}\}

x is called a generator

For example, (\mathbb{Z},+)

\langle 1 \rangle = \{+(n,1) \mid n\in \mathbb{Z}\}

But also

\langle -1 \rangle = \{+(n,-1) \mid n\in \mathbb{Z}\}

Example

We can find a generator of

(\mathbb{Z}/n\mathbb{Z},+)

Using

\langle [1] \rangle = \{[1],[2],\cdots,[n]\} = \{+(n,[1])\mid n \in \{ 1,2,\cdots,n\}\}

Corollary

(1) Assume |x| = \infty. Then H = <x^a> iff a = \pm 1

(2) Assume |x| = n < \infty then <x^a> = H iff gcd(a,n) = 1

Notice

\mathbb{Z}/6\mathbb{Z} = \left<[1]\right>,\left<[5]\right>.

Example

All the generators

(\mathbb{Z}/12\mathbb{Z},+)

Which are

\left<[1]\right>,\left<[5]\right>,\left<[7]\right>,\left<[11]\right>

Theorem

If H = \left<x\right> is a cyclic groups

(1) Every subgroup of H is cyclic

(2) If |H| = n < \infty then for each positive integer a dividing n, there is a unique subgroup of H of order \left<x^{\frac{n}{a}}\right>

Proof

(1) Let K \leq H = \left<x\right>

If K = \{1\} then done. Otherwise let a = \min\{K>0\mid x^k\in K\}.

The claim is that K is generated by \left<x^a\right>. Suppose not. Suppose there is some x^b \in K such that a\not| \,b. Thus

b = aq + r \quad 0 < r < a

Note that x^{aq} \in K. Meaning x^{-aq} \in K. Thus x^{b-aq} \in K. Thus x^r \in K. But because r is less than a, and we assumed a to be the smallest power of x in K, we have a contradiction. Thus every subgroup of H is cyclic.

(2) If H has a finite order n, we use our previous proposition to note that |\left<x^\frac{n}{a}\right>| = a.

Example

All the subgroups of \mathbb{Z}/12\mathbb{Z} are

Order 12: \left<[1]\right>

Order 6: \left<[2]\right>

Order 4: \left<[3]\right>

Order 3: \left<[4]\right>

Order 2: \left<[6]\right>

Order 1: \left<[0]\right>

Subsets

Cyclic groups generated by x are the smallest subgroups containing x.

Proposition

For any nonempty collection of subgroups of G, the intersection of all of them is also a subgroup of G.

Definition

If A is any subset of G, then

\left<A\right> = \bigcap \limits_{\substack{H\leq G\\ A \subseteq H}} H

I.e, the intersection of all subgroups containing A. Thus \left<A\right> is the minimal subgroup of G containing A.

Another way to define \left<A\right> is by using generators. Say

\overline{A} = \left\{ a_1^{\varepsilon_1},a_2^{\varepsilon_2},\cdots,a_n^{\varepsilon_n} \mid n + \mathbb{Z},\:n\geq0,\:a_1\in A,\: \varepsilon_i = \pm 1\right\}

Thus \overline{A} is the set of finite products of elements of A and inverses of these elements.

Proposition

Our definitions are equal. Symbollically \overline{A} = \left< A \right>

Proof

First we prove that \overline{A} is a group. Then

\overline{A} \not= \varnothing because A always has the identity.

if a,b \in A, then

\begin{align*} a &= a_1^{\varepsilon_1}a_2^{\varepsilon_2}\cdots a_n^{\varepsilon_n}\\ b &= b_1^{\delta_1}b_2^{\delta_2}\cdots a_n^{\delta_n} \end{align*}

then

a(b^{-1}) = a_1^{\varepsilon_1}a_2^{\varepsilon_2}\cdots a_n^{\varepsilon_n}a_n^{\delta_n}\cdots b_2^{\delta_2}b_1^{\delta_1} \in A

Thus

\overline{A} \leq G

So \left<A\right> \subseteq \overline{A}.

Now we must show \overline{A} \subseteq \left<A\right>. This is true because \left<A\right> is closed under multiplication and inverses.

Example

The subgroup

\left<(12),(13)(24)\right> \leq S_4

Fun fact, (it’s not) the subset is isomorphic to D_8

Definition

Let (G,\star) be a group. A subgroup of (G,\star) is a group (H,\star) such that H \subseteq G.

If (H,\star) is a subgroup of (G,\star), we write

(H,\star) \leq (G,\star)