Subgroup

Let $G$ be a group. The subset $H$ of $G$ is a subgroup of $G$ if $H$ is nonempty and $H$ is closed under products and inverses. (i.e., $x,y \in H \implies x^{-1} \in H \land xy \in H$). If $H$ is a subgroup of $G$, we write $H \leq G$.

Proposition 1. The Subgroup Criterion

A subset $H$ of a group $G$ is a subgroup if and only if $H \ne \emptyset$ and $\forall x,y \in H, xy^{-1} \in H$.

Proof

Suppose $H$ is a subgroup of $G$. Then, by definition of subgroup, $H \ne \emptyset$. Since $H$ is closed under products and inverses, $y \in H \implies y^{-1} \in H$ and $x,y\in H \implies xy^{-1} \in H$.

Conversely, suppose $H \ne \emptyset$ and $\forall x,y \in H, xy^{-1} \in H$.

$1 \in H$ because for any $x \in H$, $xx^{-1} = 1 \in H$. Therefore, for any $x \in H$, $1x^{-1} = x^{-1} \in H$, so $H$ is closed under inverses. Since $H$ is closed under inverses, $$ x,y \in H \implies x,y^{-1} \in H \implies x(y^{-1})^{-1} = xy \in H $$ and thus it’s also closed under products.

Therefore, by definition of subgroup, $H$ is a subgroup.

$$\tag*{$\blacksquare$}$$

Centralizer

Let $A$ be a nonempty subset of $G$. $C_G(A) = \{g \in G \mid gag^{-1} = a \text{ for all } a \in A\}$ is a subset of $G$ and called the centralizer of $A$ in $G$. Since $gag^{-1} = a$ if and only if $ga = ag$, $C_G(A)$ is the set of elements of $G$ which commute with every element of $A$.

$C_G(A)$ is a subgroup of $G$

Proof

$1 \in C_G(A)$, so $C_G(A) \ne \emptyset$. Note $$ x \in C_G(A) \implies xax^{-1} = a \text { for all } a \in A $$ and $$ y \in C_G(A) \implies yay^{-1} = a \text { for all } a \in A \implies a = y^{-1}ay\ \text{ for all } a \in A. $$

Also note that $(xy^{-1})(yx^{-1}) = 1$ and thus $(xy^{-1})^{-1} = yx^{-1}$.

Consequently, for any $a \in A$,

$$ x,y \in C_G(A) \implies (xy^{-1})a(xy^{-1})^{-1} = xy^{-1}ayx^{-1} = a \implies xy^{-1} \in C_G(A). $$

Therefore, by The Subgroup Criterion, $C_G(A)$ is a subgroup of $G$.

$$\tag*{$\blacksquare$}$$

Center

Define $Z(G) = \{g \in G \mid gx = xg \text{ for all } x \in G\}$, the set of elements commuting with all the elements of $G$. This subset of $G$ is called the center of $G$.

Note $Z(G) = C_G(G)$.

$Z(G)$ is a subgroup of $G$

Proof

Note $Z(G) = C_G(G)$. We already showed that $C_G(A)$ in general is a subgroup of $G$. $$\tag*{$\blacksquare$}$$

Normalizer

Define $gAg^{-1} = \{gag^{-1} \mid a \in A\}$. The normalizer of $A$ in $G$ is the set $$N_G(A) = \{g \in G \mid gAg^{-1} = A\}$$

Note $C_G(A) \le N_G(A)$.

$N_G(A)$ is a subgroup of $G$

Proof

$N_G(A) \ne \emptyset$ because $1 \in C_G(A) \le N_G(A)$. Now it remains to show $x,y \in N_G(A) \implies xy^{-1} \in N_G(A)$.

Suppose $x,y \in N_G(A)$. Then $\{yay^{-1} \mid a \in A\} = A$. In other words, $\varphi: A \rightarrow A$ defined by $\varphi(a) = yay^{-1}$ is a bijection because the the map is surjective and the cardinality of the domain is equal to that of the codomain. Notice $\psi: A \rightarrow A$ defined by $\psi(a) = y^{-1}ay$ is the inverse of $\varphi$ because for any $a \in A$,

$$ \varphi\psi(a) = \varphi(y^{-1}ay) = y(y^{-1}ay)y^{-1} = a. $$

Therefore, $\psi$ is also a bijection and $\{y^{-1}ay \mid a \in A\} = A$. Thus, $y^{-1} \in N_G(A)$.

Also note that $(xy^{-1})(yx^{-1}) = 1$ and thus $(xy^{-1})^{-1} = yx^{-1}$. Consequently,

$$ (xy^{-1})A(xy^{-1})^{-1} = xy^{-1}Ayx^{-1}\\ = \{x(y^{-1}ay)x^{-1} \mid a \in A\} = \{xa_yx^{-1} \mid a_y \in A\} = \{a_x \mid a_x \in A\} = A. $$

Therefore, $xy^{-1} \in N_G(A)$. By The Subgroup Criterion, $N_G(A)$ is a subgroup of $G$.

$$\tag*{$\blacksquare$}$$

Stabilizer

Let $G$ be a group acting on a set $S$, and let $s$ be some fixed element of $S$. The stabilizer of $s$ in $G$ is the set $$ G_s = \{g \in G \mid g \cdot s = s\} $$

$G_s$ is a subgroup of $G$

Proof

By definition of $G_s$, $G_s$ is a subset of $G$. Suppose $y\in G_s$. Then, it follows from the definition of group action and $y \cdot s = s$ that $$ y^{-1}\cdot s = y^{-1} \cdot (y \cdot s) = (y^{-1}y)\cdot s = 1 \cdot s = s. $$

Further suppose that $x \in G_s$. Then, $$ xy \cdot s = xy\cdot (y^{-1} \cdot s) = (xyy^{-1})\cdot s = x \cdot s = s. $$

Therefore, $G_s$ is closed under products and inverses. By definition of subgroup, $G_s$ is a subgroup of $G$.

$$\tag*{$\blacksquare$}$$

Kernel

The kernel of an action of $G$ on $S$ is defined as $\{g \in G \mid g\cdot s = s \text{ for all } s \in S\}$.

The kernel of an action is a subgroup

Proof

Let $K$ be the kernel. By its definition, $K$ is a subset of $G$. Suppose $y \in K$. Then $y \cdot s = s$ for all $s \in S$. It follows that, for an arbitrary $s \in S$, $$ y^{-1} \cdot s = y^{-1}\cdot (y\cdot s) = (y^{-1}y)\cdot s = 1 \cdot s = s. $$

Further suppose that $x \in K$. Then, $$ xy\cdot s = xy\cdot(y^{-1}\cdot s) = (xyy^{-1})\cdot s = x \cdot s = s. $$

Therefore, $K$ is closed under products and inverses. By definition of subgroup, $K$ is a subgroup of $G$.

$$\tag*{$\blacksquare$}$$

Cyclic Group

A group $H$ is cyclic if there is some element $x \in H$ such that $H = \{x^n \mid n\in\mathbb{Z}\}$. We write $H = \langle x \rangle$ and say $H$ is generated by $x$, and $x$ is a generator of $H$.

Proposition 2

Let $H = \langle x \rangle$.

if $|H| = n \lt \infty $, then $x^n = 1$ and $1,x,x^2,…x^{n-1}$ are all distinct elements of $H$
if $|H| = \infty $, then $x^n \ne 1$ for all $n \ne 0$ and $x^a \ne x^b$ for all $a \ne b$ in $\mathbb{Z}$.

Proof

For the first part, suppose $|H| = n$ and let $k$ be the order of $x$.

Using the Division Algorithm, any $i \in \mathbb{Z}$ can be written as $i = kq + r$. Therefore, for an arbitrary $i \in \mathbb{Z}$, $x^i = x^{kq + r} = x^{kq}x^r = 1^qx^r = x^r$. Since $r \in [0, k-1]$, $H = \{1, x, x^2,…,x^{k - 1}\}$.

Let $0 \le a \le b \lt k$. Then $x^a = x^b \implies 1 = x^{b-a}$. Since identity is unique in a group and we know that one identity $x^0 = 1$ exists already, $a$ must equal $b$. Therefore, $a \ne b \implies x^a \ne x^b$.

It follows that all the elements in $H = \{1,x,x^2,…,x^{k-1}\}$ shown above are distinct. Since it’s given that $|H| = n$, $k = n$. Hence, $x^n = 1$ and $1,x,x^2,…,x^{n-1}$ are all distinct elements of $H$.

For the second part, suppose $|H| = \infty$.

Suppose for a contradiction that $x^n = 1$ for some $n \ne 0$. Then, using the Division Algorithm, for any $i \in \mathbb{Z}$, $x^i = x^{nq+r} = x^{nq}x^r = x^r$. Since $r \in [0, n-1]$, $H = \{1,x,x^2,…,x^{n-1}\}$, contradicting the fact that $|H| = \infty$. Therefore, $x^n \ne 1$ for all $n \ne 0$.

Suppose for a contradiction that $x^a = x^b$ for some $a \ne b \in \mathbb{Z}$. Then $1 = x^{b-a}$. Because of the uniqueness of identity, $a = b$, contradicting our assumption. Therefore, $x^a \ne x^b$ for all $a \ne b$ in $\mathbb{Z}$.

$$\tag*{$\blacksquare$}$$

Proposition 3

Let $G$ be a group, $x \in G$, and $m, n \in \mathbb{Z}$. If $x^n = 1$ and $x^m = 1$, then $x^d = 1$, where $d = \gcd(m,n)$. In particular, if $x^m = 1$ for some $m \in \mathbb{Z}$, then $|x| \mid m$.

Proof

By the Euclidean Algorithm, there exist $r, s \in \mathbb{Z}$ such that $d = mr + ns$. Therefore, $$ x^d = x^{mr + ns} = x^{mr}x^{ns} = 1^r1^s = 1. $$

For the second part, let $x^m = 1$ for some $m \in \mathbb{Z}$ and let $n = |x|$. Then since $x^m = x^n = 1$, using what we’ve just shown, $x^d = 1$ where $d = \gcd(m, n)$. Note that $0 \lt d \le n$ and that, by definition of order, $n$ is the smallest positive power of $x$ to give the identity. Therefore, $n = d = \gcd(m,n)$, and thus $n \mid m$.

$$\tag*{$\blacksquare$}$$

Theorem 4

If $\langle x \rangle$ and $\langle y \rangle$ are both cyclic groups of order $n$, the map $\varphi: \langle x \rangle \rightarrow \langle y \rangle$ defined by $x^k \mapsto y^k$ is well defined and is an isomorphism.
If $\langle x \rangle$ is an infinite cyclic group, the map $\varphi: \mathbb{Z} \rightarrow \langle x \rangle$ defined by $k \mapsto x^k$ is well defined and is an isomorphism.

Proof

To show that the map is well defined, we need to show that $s = t \implies \varphi(s) = \varphi(t)$.

(1)

Suppose $x^a = x^b$. Since $x^{a-b} = 1$, as shown earlier, $n \mid a-b$. Then we can write $a-b = tn$ or $a = tn + b$ for some $t$. It follows that $\varphi(x^a) = y^a = y^{tn+b} = y^{tn}y^b = y^b = \varphi(x^b)$. Therefore, $\varphi$ is well defined.

For arbitrary $i,j \in \mathbb{Z}$, $\varphi(x^i)\varphi(x^j) = y^iy^j = y^{i+j} = \varphi(x^{i+j}) = \varphi(x^ix^j)$. Thus, $\varphi$ is a homomorphism.

Since $\varphi$ is both a homomorphism and a bijection, it’s an isomorphism.

(2)

$a = b \implies \varphi(a) = x^a = x^b = \varphi(b)$. Therefore, $\varphi$ is well defined.

Note $\mathbb{Z}$ is a group under addition. For arbitrary $i,j \in \mathbb{Z}$, $\varphi(i)\varphi(j) = x^ix^j = x^{i+j} = \varphi(i+j)$. Therefore, $\varphi$ is a homomorphism.

For any $x^k \in \langle x \rangle$, $\varphi(k) = x^k$, so $\varphi$ is surjective. We’ve shown above that if $\langle x \rangle$ is an infinite group, then $x^a \ne x^b$ for all $a \ne b \in \mathbb{Z}$. Therefore, $\varphi$ is injective.

Since $\varphi$ is both a homomorphism and a bijection, it’s an isomorphism.

$$\tag*{$\blacksquare$}$$

Proposition 5

Let $G$ be a group, $x \in G$, and $a \in \mathbb{Z} - \{0\}$.

If $|x| = \infty$, then $|x^a| = \infty$.
If $|x| = n < \infty$, then $|x^a| = \frac{n}{\gcd(n,a)}$.
In particular, if $|x| = n < \infty$ and $a$ is a positive integer dividing $n$, then $|x^a| = \frac{n}{a}$.

Proof

(1)

Suppose $|x^a| = n < \infty$. Then $(x^a)^n = x^{an} = 1$. Also, $x^{-an} = (x^{an})^{-1} = 1^{-1} = 1$. By definition of order, $n \ne 0$, and $a \ne 0$ as given. Therefore, either $an$ or $-an$ is positive. Thefore, $|x| \ne \infty$.

That is, $|x^a| \ne \infty \implies |x| \ne \infty$. Equivalently, $|x| = \infty \implies |x^a| = \infty$.

(2)

Let $d = \gcd(n,a)$. Then we can write $n = db$ and $a = dc$ for some $b,c \in \mathbb{Z}$ where $b$ and $c$ are relatively prime and $b > 0$. We want to show that $|x^a| = b$.

Let $k = |x^a|$. Then $1 = (x^a)^k = x^{ak}$, and thus, as shown earlier, $n \mid ak$. Equivalently, $db \mid dck$, or $b \mid ck$. Since $b$ and $c$ are relatively prime, $b \mid k$.

Also, $(x^a)^b = x^{ab} = x^{dcb} = x^{nc} = (x^n)^c = 1^c = 1$, and thus $k \mid b$.

We just showed that $k$ and $b$ are positive integers that divide each other. Therefore, $k = b$, or $|x^a| = b$.

$$\tag*{$\blacksquare$}$$

(3)

TODO

Proposition 6

Let $H = \langle x \rangle$.

Assume $|x| = \infty$. Then $H = \langle x^a \rangle$ if and only if $a = \pm1$.
Assume $|x| = n < \infty$. Then $H = \langle x^a \rangle$ if and only if $\gcd(a,n) = 1$.

Proof

(1)

Suppose $H = \langle x \rangle = \langle x^a \rangle$. Since $x \in \langle x \rangle$, $x \in \langle x^a \rangle$ too. Therefore, there exists some $k \in \mathbb{Z}$ such that $x = (x^a)^k = x^{ak}$. Then $1 = x^{ak-1}$. Since $|x| = \infty$, $|H| = \infty$, and as shown earlier, $x^s \ne x^t$ for all $s \ne t \in \mathbb{Z}$. Therefore, $ak-1$ must be $0$. Since $a,k \in \mathbb{Z}$, $a = \pm 1$.

Conversely, suppose $a = \pm 1$. If $a = 1$, $\langle x^a \rangle = \langle x^1 \rangle = \langle x \rangle$. Now suppose $a = -1$. Then for any element $x^i \in \langle x \rangle$, $x^i = (x^{-1})^{-i} = (x^a)^{-i} \in \langle x^a \rangle$. Similarly, for any element $(x^a)^i \in \langle x^a \rangle$, $(x^a)^i = x^{-i} \in \langle x \rangle$. Therefore, $\langle x \rangle = \langle x^a \rangle$.

(2)

$\langle x^a \rangle$ is a finite subgroup of $\langle x \rangle$. We showed earlier that $|x^a| = \frac{n}{\gcd(n,a)}$ where $n = |x|$. Thus, $\langle x \rangle = \langle x^a \rangle$ iff $|\langle x \rangle| = |\langle x^a \rangle|$ iff $n = \frac{n}{\gcd(n,a)}$ iff $\gcd(n,a) = 1$.

$$\tag*{$\blacksquare$}$$

Theorem 7.1

Let $H = \langle x \rangle$ be a cyclic group. If $K \le H$, then either $K = \{1\}$ or $K = \langle x^d \rangle$, where $d$ is the smallest positive integer such that $x^d \in K$.

Proof

Suppose $K \le H$. $\{1\} \le H$, so $K$ can be $\{1\}$.

Now consider the case where $K \ne \{1\}$. Note all non-identity elements of $K$ can be expressed as a nonzero power of $x$. In particular, there is at least one element $x^k$ in $K$ for a nonzero integer $k$. Since $x^k \in K \implies x^{-k} \in K$, there is at least one positive power of $x$ and one negative power of $x$ in $K$.

Let $d$ be the smallest positive integer such that $x^d \in K$. Then, because $K$ is a group and thus closed, $(x^d)^n \in K$ for all $n \in \mathbb{Z}$. Therefore, $\langle x^d \rangle \le K$.

Suppose for a contradiction that there exists $j$ such that $x^j \in K \land x^j \notin \langle x^d \rangle$. As argued above, we can assume $j > 0$ without loss of generality. Furthermore, by definition of $d$, $j > d$. Since $x^j \notin \langle x^d \rangle$, $j$ is not a multiple of $d$. Therefore, using the Division Algorithm, we can write $j = qd + r$ where $q > 0$ and $d > r > 0$. Since $x^{qd} = (x^d)^q \in K$, its inverse is also in $K$. It follows that $x^j \in K \implies x^{qd+r} \in K = x^{qd}x^r \in K \implies x^r \in K$.

However, this contradicts the fact that $d$ is the smallest positive integer such that $x^d \in K$. Therefore, there is no $j$ such that $x^j \in K \land x^j \notin \langle x^d \rangle$. Hence, $K \le \langle x^d \rangle$.

Since $\langle x^d \rangle \le K$ and $K \le \langle x^d \rangle$, $K = \langle x^d \rangle$.

$$\tag*{$\blacksquare$}$$

Theorem 7.2

Let $H = \langle x \rangle$ be a cyclic group. If $|H| = \infty$, then for any distinct nonnegative integers $a$ and $b$, $\langle x^a \rangle \ne \langle x^b \rangle$. Furthermore, for every $m \in \mathbb{Z}$, $\langle x^m \rangle = \langle x^{|m|} \rangle$, where $|m|$ denotes the absolute value of $m$, so that the nontrivial subgroups of $H$ correspond bijectively with the integers $1,2,3…$.

Proof

Suppose $a = 0$ and $b \ne 0$. Then $\langle x^a \rangle = \{1\}$, and $\langle x^b \rangle \ne \{1\}$. Therefore, $\langle x^a \rangle \ne \langle x^b \rangle$. The same logic applies when $a \ne 0$ and $b = 0$.

Now assume $a$ and $b$ are both nonzero. Since $\langle x^a \rangle$ is a subgroup of $H$, as shown previously, we can let $\langle x^a \rangle = \langle x^\alpha \rangle$ where $\alpha$ is the smallest positive integer such that $x^\alpha \in \langle x^a \rangle$. Likewise, let $\langle x^b \rangle = \langle x^\beta \rangle$ where $\beta$ is the smallest positive integer such that $x^\beta \in \langle x^b \rangle$.

Without loss of generality, assume $\alpha < \beta$. Suppose for a contradiction that $\langle x^a \rangle = \langle x^b \rangle$. Then $x^\alpha \in \langle x^b \rangle$. Since $\alpha < \beta$, this contradicts the fact that $\beta$ is the smallest positive integer such that $x^\beta \in \langle x^b \rangle$. Therefore, $\langle x^a \rangle \ne \langle x^b \rangle$.

The second assertion is trivially true if $m \ge 0$, so assume $m < 0$. We want to show that $\langle x^m \rangle = \langle x^{-m} \rangle$.

For any $(x^m)^k \in \langle x^m \rangle$, $(x^m)^k = (x^{-m})^{-k} \in \langle x^{-m} \rangle$. Conversely, for any $(x^{-m})^k \in \langle x^{-m} \rangle$, $(x^{-m})^k = (x^m)^{-k} \in \langle x^m \rangle$. Therefore, $\langle x^m \rangle = \langle x^{-m} \rangle$.

$$\tag*{$\blacksquare$}$$

Theorem 7.3

Let $H = \langle x \rangle$ be a cyclic group. If $|H| = n < \infty$, then for each positive integer $a$ dividing $n$ there is a unique subgroup of $H$ of order $a$. This subgroup is the cyclic group $\langle x^d \rangle$ where $d = {\frac{n}{a}}$. Furthermore, for every $m \in \mathbb{Z}$, $\langle x^m \rangle = \langle x^{\gcd(n,m)} \rangle$, so that the subgroups of $H$ correspond bijectively with the positive divisors of $n$.

Proof

Let $a \mid n$ and $n = ad$. Then $d = \frac{n}{a}$ and $a = \frac{n}{d}$. By Proposition 5, $$ |x| = n \implies |x^a| = \frac{n}{\gcd(n,a)} $$

Applying this, we have $|x^d| = \frac{n}{\gcd(n,d)} = \frac{n}{d} = a$. Therefore, $\langle x^d \rangle$ is a subgroup of $H$ of order $a$.

To show uniqueness, suppose $K$ is any subgroup of order $a$. Then we can let $K = \langle x^b \rangle$ where $b$ is the smallest positive integer such that $x^b \in K$. Then, again using Proposition 5,

$$ \frac{n}{d} = a = |K| = |\langle x^b \rangle| = \frac{n}{\gcd(n,b)} $$

Therefore, $d = \gcd(n,b)$ and in particular, $d \mid b$. Since $b$ is a multiple of $d$, $x^b \in \langle x^d \rangle$. That is, $K = \langle x^b \rangle \le \langle x^d \rangle$. Since $|\langle x^b \rangle| = |\langle x^d \rangle| = a$, $K = \langle x^b \rangle = \langle x^d \rangle$.

Now consider the second assertion. Since $\gcd(n,m) \mid m$, any $x^{mk} \in \langle x^m \rangle$ also belongs to $\langle x^{\gcd(n,m)} \rangle$. Therefore, $\langle x^m \rangle \le \langle x^{\gcd(n,m)} \rangle$. Note, using Proposition 5 again,

$$ |\langle x^m \rangle| = \frac{n}{\gcd(n,m)} = \frac{n}{\gcd(n,\gcd(n,m))} = |\langle x^{\gcd(n,m)} \rangle|. $$

Therefore, $\langle x^m \rangle = \langle x^{\gcd(n,m)} \rangle$.

$$\tag*{$\blacksquare$}$$

Subgroup

Proposition 1. The Subgroup Criterion

Proof

Centralizer

\(C_G(A)\) is a subgroup of \(G\)

Proof

Center

\(Z(G)\) is a subgroup of \(G\)

Proof

Normalizer

\(N_G(A)\) is a subgroup of \(G\)

Proof

Stabilizer

\(G_s\) is a subgroup of \(G\)

Proof

Kernel

The kernel of an action is a subgroup

Proof

Cyclic Group

Proposition 2

Proof

Proposition 3

Proof

Theorem 4

Proof

(1)

(2)

Proposition 5

Proof

(1)

(2)

(3)

Proposition 6

Proof

(1)

(2)

Theorem 7.1

Proof

Theorem 7.2

Proof

Theorem 7.3

Proof