In this lecture, we study maps from a group into another group that carry over the same group structure i.e preserve group operations. Such maps are called homomorphisms. It turns out that it suffices to require a homomorphism to preserve a binary operation only. Then the unary operation (inverses) and the nullary operation (the identity element) are automatically preserved.

Definition. Let $(G,\ast)$ and $(G’,\sharp)$ be two groups. A map $\varphi: G\longrightarrow G’$ is called a homomorphism if $\varphi(a\ast b)=\varphi(a)\sharp\varphi(b)$.

Example. Define $\varphi: (\mathbb{R}^+,\cdot)\longrightarrow(\mathbb{R},+)$ by
$$\varphi(x)=\log_{10}x,\ \forall x\in\mathbb{R}^+.$$
Then $\varphi$ is a homomorphism. In addition, $\varphi$ is one-to-one and onto. A homomorphism which is one-to-one and onto is called an isomorphism. If there is an isomorphism from a group onto another group, they are the same group meaning we do not distinguish them.

Example. The map $\varphi: \mathbb{Z}\longrightarrow\mathbb{Z}_n$ defined by
$$\varphi(x)=[x],\ \forall\ x\in\mathbb{Z}$$
is a homomorphism called the canonical or the natural homomorphism. The canonical homomorphism is onto.

Example. Let $G$ be a group and $A(G)$ the set of all bijective (i.e. one-to-one and onto) maps of $G$ onto itself. Recall that $A(G)$ forms a group with composition. Let $a\in G$. Define $T_a: G\longrightarrow G$ by
$$T_ax=ax\ \forall\ x\in G.$$ Hence $\forall a\in G$, $T_a\in A(G)$. Furthermore, $\forall a,b\in G$, $T_aT_b=T_{ab}$: $\forall x\in G$,
$$T_aT_b(x)=T_a(T_bx)=T_a(bx)=a(bx)=(ab)x=T_{ab}x.$$
Define $\varphi: G\longrightarrow A(G)$ by
$$\varphi(a)=T_a,\ \forall\ a\in G.$$ Then $\varphi$ is a homomorphism. In addition, it is one-to-one. If $|G|=n$, then $|A(G)|=n!$, so $\varphi$ cannot be onto.

Definition. A homomorphism which is one-to-one (injective) is called a monomorphism. A homomorphism which is onto (surjective) is called an epimorphism. A homomorphism which is one-to-one and onto (bijective) is called an isomorphism as mentioned above. $G$ is said to be isomorphic to $G’$ if there exists an isomorphism from $G$ onto $G’$. If $G$ is isomorphic to $G’$, we write $G\cong G’$. An isomorphism from a group $G$ onto itself is called an automorphism.

Theorem. [Cayley’s Theorem] Every group $G$ is isomorphic to some subgroup of $A(S)$, for an appropriate set $S$.

We in fact proved this theorem in the previous example by taking $S$ to be the group $G$ itself. But there may be other choices for $S$. If $G$ is finite, $S$ may be taken to be a finite set in which case $A(S)$ is $S_n$, the group of permutations of $\{1,2,\cdots,n\}$. In this case, Cayley’s Theorem is stated as: A finite group can be represented as a group of permutations.

Example. Let $G$ be a group and $a\in G$ be fixed. The map $\varphi: G\longrightarrow G$ defined by
$$\varphi(x)=a^{-1}xa,\ \forall x\in G$$
is an automorphism called the inner automorphism of $G$ induced by $a$.

Example. Define $\varphi: (\mathbb{R},+)\longrightarrow (\mathbb{C}\setminus\{0\},\cdot)$ by
$$\varphi(x)=e^{ix},\ \forall x\in\mathbb{R}.$$
While $\varphi$ is a homomorphism, it is neither one-to-one nor onto.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then

(a) $\varphi(e)=e’$.

(b) $\varphi(a^{-1})=\varphi(a)^{-1}$.

Proof. (a) Let $a\in G$. Then $ae=a$. Since $\varphi$ is a homomorphism, $$\varphi(ae)=\varphi(a)\varphi(e)=\varphi(a)=\varphi(a)e’.$$ So by the cancellation law, we obatin $\varphi(e)=e’$.

(b) Let $a\in G$. Then
\begin{align*}
\varphi(a)\varphi(a^{-1})&=\varphi(aa^{-1})\ (\mbox{because $\varphi$ is a homomorphism})\\
&=\varphi(e)\\
&=e’\ (\mbox{by part (a)})
\end{align*}
Hence, $\varphi(a)^{-1}=\varphi(a^{-1})$.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\varphi(G)\leq G’$.

In here, the kernel of a function was introduced. Let $\varphi: G\longrightarrow G’$ be a homomorphism and $K=\varphi^{-1}(e’)=\{a\in G: \varphi(a)=e’\}$. Then $\ker\varphi$ and $K=\varphi^{-1}(e’)$, the preimage of $e’$ under $\varphi$ are closely related as:
\begin{align*}
(a,b)\in\ker\varphi&\Longleftrightarrow\varphi(a)=\varphi(b)\\
&\Longleftrightarrow\varphi(ab^{-1})=e’\\
&\Longleftrightarrow ab^{-1}\in K.
\end{align*}
So, in group theory we define $K=\varphi^{-1}(e’)$ to be $\ker\varphi$, the kernel of $\varphi$.

Theorem. $\varphi: G\longrightarrow G’$ is one-to-one if and only if $\ker\varphi=\{e\}$.

Proof. ($\Rightarrow$) Suppose that $\varphi$ is one-to-one and let $a\in\ker\varphi$. Then $\varphi(a)=\varphi(e)=e’$. Since $\varphi$ is one-to-one, $a=e$.

($\Leftarrow$) Suppose that $\ker\varphi=\{e\}$ and let $\varphi(a)=\varphi(b)$. Then
\begin{align*}
\varphi(ab^{-1})=e’&\Longrightarrow ab^{-1}\in\ker\varphi\\
&\Longrightarrow ab^{-1}=e\\
&\Longrightarrow a=b.
\end{align*}
So, $\varphi$ is one-to-one.

Theorem. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\leq G$.

Theorem. Let $K=\ker\varphi$. Then $\forall a\in G$, $aK=Ka$.

Proof. It suffices to show (why?) that $\forall a\in G$, $a^{-1}Ka\subset K$. Let $a\in G$. Then $\forall k\in K$,
$$\varphi(a^{-1}ka)=\varphi(a)^{-1}\varphi(k)\varphi(a)=e’\Longrightarrow a^{-1}ka\in K.$$

In this lecture, we study the congruence modulo $n$, $\equiv\mod n$ on $\mathbb{Z}$. As we shall see, $\equiv\mod n$ is an equivalence relation on $\mathbb{Z}$. It is indeed more than just an equivalence relation. $\equiv\mod n$ also preserves operations $+$ and $\cdot$ on $\mathbb{Z}$. Because of this, we can define an operation $+$ on the quotient set $\mathbb{Z}/\equiv\mod n$ so that it becomes a quotient group. In general, if an equivalence relation preserves operations, it is called a congruence relation and a congruence relation allows us to define a quotient algebra e.g. a quotient group. Let $G$ be a group and $H\leq G$. We studied the equivalence relation $\sim$ on $G$: $\forall a,b\in G$,
$$a\sim b \Longleftrightarrow ab^{-1}\in H.$$ It turns out that $\sim$ is a congruence relation i.e. it preserves the group operation $\cdot$ if $H$ is a normal subgroup of $G$. I am not going to talk about normal subgroups here but we will study details about this next time.

On $\mathbb{Z}$, define $a\equiv b\mod n$, we say $a$ is congruent to $b$ modulo $n$, if $n|b-a$. The class of $a$
$$[a]=\{a+nk: k\in\mathbb{Z}\}$$ is called the congruence class of $a$. By Euclid’s algorithm, $\forall b\in\mathbb{Z}$, $b=qn+r$ where $0\leq r<n$. This implies that $b\equiv r\mod n$and so $[b]=[r]$. Hence, the $n$ classes $[0],[1],\cdots,[n-1]$ give us all the congruence classes and they are all distinct. Let
$$\mathbb{Z}_n=\{[0],[1],\cdots,[n-1]\}.$$ That is, $\mathbb{Z}_n$ is the quotient set modulo the equivalence relation $\equiv\mod n$, $\mathbb{Z}/\equiv\mod n$. Define $+$ on $\mathbb{Z}_n$: $\forall [a],[b]\in\mathbb{Z}_n$,
$$[a]+[b]:=[a+b].$$ Then $+$ is well-defined on $\mathbb{Z}_n$. What this means is that $+$ is well-defined as a function $+:\mathbb{Z}_n\times\mathbb{Z}_n\longrightarrow\mathbb{Z}_n$. In other words, if $[a]=[a’]$ and $[b]=[b’]$ then $[a+b]=[a’+b’]$. This can be shown straightforward and is left for the readers. One can also readily see that $(\mathbb{Z}_n,+)$ is an abelian group. Define $\cdot $ on $\mathbb{Z}_n$: $\forall [a],[b]\in\mathbb{Z}_n$,
$$[a]\cdot[b]:=[ab].$$ Then $\cdot$ is also well-defined on $\mathbb{Z}_n$. However, $(\mathbb{Z}_n,\cdot)$ is not a group. $\forall [a]\in\mathbb{Z}_n$, $[0][a]=[0]$ and $[1]$ is the identity under multiplication, so $[0]$ does not have a multiplicative inverse. Would nonzero elements of $\mathbb{Z}_n$ form a group then? The answer is no. For instance, in $\mathbb{Z}_6$, $[2],[3]\ne 0$ but $[2][3]=[6]=[0]$. So, is there a subset of $\mathbb{Z}_n$ that forms a group under $\cdot$? The answer is affirmative. Let $\mathbb{Z}_n^\ast=\{[a]\in\mathbb{Z}_n: (a,n)=1\}$. First, note that for $[a]=[b]$, $(a,n)=1$ if and only if $(b,n)=1$. Recall the property that if $(a,n)=1$ and $(b,n)=1$, then $(ab,n)=1$. This implies that $[a][b]=[ab]\in\mathbb{Z}_n^\ast$. Now suppose that $[a][b]=[a][c]$. Then
\begin{align*}
[ab]=[ac]&\Longrightarrow ab-ac=nk\\
&\Longrightarrow n|a(b-c)\\
&\Longrightarrow n|b-c\ (\mbox{since}\ (a,n)=1)\\
&\Longrightarrow [b]=[c].
\end{align*}
That is, $\mathbb{Z}_n^\ast$ has the cancellation property, so $(\mathbb{Z}_n^\ast,\cdot)$ is a group. In some textbooks, $\mathbb{Z}_n^\ast$ is also denoted by $U_n$. The order of $\mathbb{Z}_n^\ast$ is the number of integers $1\leq m<n$ such that $(m,n)=1$.

Definition. The Euler $\varphi$-function, $\varphi(n)$, is defined by $\varphi(1)=1$ and for $n>1$, $\varphi(n)=$ the number of positive integers $m$ with $1\leq m<n$ such that $(m,n)=1$. Hence, $|\mathbb{Z}_n^\ast|=\varphi(n)$. If $n=p$, a prime, then $\varphi(p)=p-1$.

Note that the Euler $\varphi$-function is multiplicative, namely if $(m,n)=1$, then $\varphi(mn)=\varphi(m)\varphi(n)$.

Example. $\varphi(8)=4$, $\varphi(15)=8$.

Example.
\begin{align*}
\mathbb{Z}_8^\ast&=\{[1],[3],[5],[7]\},\\
\mathbb{Z}_{15}^\ast&=\{[1],[2],[4],[7],[8],[11],[13],[14]\}.
\end{align*}

Example. $\mathbb{Z}_9^\ast=\{[1],[2],[4],[5],[7],[8]\}$.
\begin{align*}
[1]^1&=[2],\ [2]^2=[4],\ [2]^3=[8],\ [2]^4=[16]=[7],\\
[2]^5&=[32]=[5],\ [2]^6=[2][2]^5=[2][5]=[10]=[1].
\end{align*}
Hence, $\mathbb{Z}_9^\ast$ is a cyclic group $\mathbb{Z}_9^\ast=\langle[2]\rangle$.

As a summary,

Theorem. $\mathbb{Z}_n^\ast$ forms an abelian group under the product $[a][b]=[ab]$, of order $\varphi(n)$, where $\varphi(n)$ is the Euler $\varphi$-function.

Theorem. [Euler] If $(a,n)=1$ then $a^{\varphi(n)}\equiv 1\mod n$.

Proof. Since $\mathbb{Z}_n^\ast$ is a group of order $\varphi(n)$, $a^{\varphi(n)}=1$, $\forall a\in\mathbb{Z}_n^\ast$. So,
$$[a^{\varphi(n)}]=[a]^{\varphi(n)}=[1]\Longrightarrow a^{\varphi(n)}\equiv 1\mod n.$$

Corollary. [Fermat] If $p$ is a prime and $p\not | a$, then
$$a^{p-1}\equiv 1\mod p.$$
For any integer $b$, $b^p\equiv b\mod p$. This corollary is usually called Fermat’s Little Theorem.

Proof. $\varphi(p)=p-1$. If $(a,p)=1$, then by the previous theorem,
\begin{align*}
a^{p-1}\equiv 1\mod p &\Longrightarrow a^{p-1}-1=pk\\
&\Longrightarrow a^p-a=p(ak)\\
&\Longrightarrow a^p\equiv a\mod p.
\end{align*}
If $p\not| b$, then $b^p\equiv b\mod p$. If $p|b$, then $b\equiv 0\mod p\Longrightarrow b^p\equiv 0\mod p$. So for any $b\in\mathbb{Z}$, $b^p\equiv b\mod p$.

Example. Compute the remainder of $8^{103}$ when divided by 13.

Solution. Since $13\not|8$, by Fermat’s Little Theorem we have $8^{13-1}=8^{12}\equiv 1\mod 13$. Dividing 103 by 13, we obtain $103=12\cdot 8+7$. So,
\begin{align*}
8^{103}&\equiv (8^{12})^88^7\equiv 8^7\equiv (-5)^7\ (8\equiv -5\mod 13)\\
&\equiv (25)^3(-5)\equiv (-1)^3(-5)\equiv 5\mod 13\ (25\equiv -1\mod 13).
\end{align*}

Example. Show that $2^{11,213}-1$ is not divisible by 11.

Solution. Since $11\not| 2$, by Fermat’s Little Theorem we have $2^{10}\equiv 1\mod 11$. $11,213=10\cdot 1,121+3$. So,
\begin{align*}
2^{11,213}-1&\equiv (2^{10})^{1,121}\cdot 2^3-1\equiv 2^3-1\\
&\equiv 7\mod 11.
\end{align*}
Hence, when $2^{11,213}-1$ is divided by 11, the remainder is 7.

The number $11,213$ is a prime and $2^{11,213}-1$ is also a prime. In number theory, primes of the form $2^p-1$ where $p$ is a prime are called Mersenne primes.

In this lecture, we study Lagrange’s Theorem which is named after the Italian/French mathematician, physicists and astronomer Joseph-Louis Lagrange. It states that the order of a subgroup $H$ of a finite group $G$ divides the order of $G$. Lagrange’s Theorem has many important applications in group theory. Before we discuss the theorem we first need to study an important class of binary relations called equivalence relations.

Definition. A binary relation $R$ on a nonempty set $S$ is a subset $R\subset S\times S$. If $(a,b)\in R$, we say $a$ is $R$-related to $b$ and write $aRb$. A binary relation $R$ on a set $S$ is called an equivalence relation on $S$ if it satisfies:

1. $aRa$ $\forall a\in S$. ($R$ is reflexive.)

2. $\forall a,b,\in S$, $aRb\Longrightarrow bRa$. ($R$ is symmetric.)

3. $\forall a,b,c\in S$, $aRb$ and $bRc$ $\Longrightarrow$ $aRc$. ($R$ is transitive.)

Examples. 1. Define a binary relation $\equiv\mod n$ on $\mathbb{Z}$, the set of integers by
$$\forall a,b\in\mathbb{Z},\ a\equiv b\mod n\ \mbox{if}\ n|(a-b).$$ Then $\equiv\mod n$ is an equivalence relation in $\mathbb{Z}$, called the congruence relation modulo $n$. If $a\equiv b\mod n$, we say $a$ is congruent to $b$ modulo $n$. Note that $n|(a-b)$ if and only if $a-b=nk$ for some $k\in\mathbb{Z}$ if and only if $a$ and $b$ have the same remainder when they are divided by $n$ if and only if $a-b\in n\mathbb{Z}$ where
$$n\mathbb{Z}=\{nk: k\in\mathbb{Z}\},$$ the set of all integer multiples of $n$. $\forall n\in\mathbb{Z}$, $n\mathbb{Z}$ is a subgroup of the additive group $(\mathbb{Z},+)$.

2. Let $G$ be a group and $H\leq G$. Define a binary relation $R$ on $G$ by
$$\forall a,b\in G,\ aRb\ \mbox{if}\ ab^{-1}\in H.$$ Then $R$ is an equivalence relation on $G$.

Definition. If $R$ is an equivalence relation on a set $S$, then $[a]$, the equivalence class of $a\in S$, is defined by
$$[a]=\{b\in S: bRa\}.$$

Theorem. Let $R$ be an equivalence relation on a set $S$. Then

1. $S=\bigcup_{a\in S}[a]$.

2. $\forall a,b\in S$, either $[a]=[b]$ or $[a]\cap [b]=\emptyset$.

This means that an equivalence relation $R$ on a set $S$ partitions $S$ into equivalence classes. The set of all equivalence classes $\{[a]: a\in S\}$ is denoted by $S/R$ and is called the quotient set of $S$ modulo $R$.

The converse of the above theorem is also true, namely

Theorem. Let $\wp=\{A_i:i\in I\}$ be a partition of a set $S$ i.e. $S=\bigcup_{i\in I}A_i$ and $\forall i\ne j\in I$, $A_i\cap A_j=\emptyset$. Define a binary relation $R$ on $S$ by
$$\forall a,b\in S,\ aRb\ \mbox{if}\ a,b\in A_i\ \mbox{for some}\ i\in I.$$ Then $R$ is an equivalence relation on $S$, called the equivalence relation induced by the partition $\wp$.

Example. Let $f: X\longrightarrow Y$ be a surjective map. Then $\{f^{-1}(y): y\in Y\}$ forms a partition of $X$. The equivalence relation induced by this partition is
$$\ker f=\{(a,b)\in X\times X: f(a)=f(b)\}.$$ This equivalence relation is called the kernel of the function $f$.

Let $G$ be a group and $H\leq G$. Consider the equivalence relation $R$ on $G$ given by
$$\forall a,b\in G,\ aRb\ \mbox{if}\ ab^{-1}\in H.$$
Then for each $a\in G$, $[a]=Ha$ where $Ha=\{ha:h\in H\}$. $Ha$ is called a right coset of $H$ in $G$.

Example. Consider $\equiv\mod n$, the congruence relation modulo $n$ on $\mathbb{Z}$. For each $a\in \mathbb{Z}$, $[a]=n\mathbb{Z}+a$. The right coset $n\mathbb{Z}+a$ is called the congruence class of $a$ modulo $n$.

Now we are ready to discuss Lagrange’s Theorem.

Theorem [Lagrange’s Theorem] If $G$ is a finite group and $H\leq G$, then $|H|||G|$.

Proof. $G=Ha_1\cup Ha_2\cup\cdots\cup Ha_k$ and $Ha_i\cup Ha_j=\emptyset$ if $i\ne j$, $i,j=1,2,\dots, k$. Due to the cancellation law, we see that $|H|=|Ha_i|$ for all $i=1,2,\cdots, k$. Therefore,
\begin{align*}
|G|&=\sum_{i=1}^k|Ha_i|\\
&=\sum_{i=1}^k|H|\\
&=k|H|.
\end{align*}
This completes the proof.

If $G$ is finite, the number of right cosets of $H$ in $G$, namely $\frac{|G|}{|H|}$ is called the index of $H$ in $G$ and is written as $|G:H|$ or $i_G(H)$. Here, we use the notation $|G:H|$.

It should be noted that the converse of Lagrange’s Theorem need not be true. For example, the alternating group of index 4
$$A_4=\{1,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$$ has no subgroup of order 6 though $|A_4|=12$ and $6|12$.

Theorem. A group $G$ of prime order is cyclic.

Proof. Let $|G|=p$, a prime. Let $H\leq G$. Then by Lagrange’s Theorem, $|H|||G|$ and so either $|H|=1$ or $|H|=p$. If $H\ne\{e\}$, then $H=G$. If $a\ne e\in G$, $\langle a\rangle=G$.

Let $G$ be a finite group and $a\ne e\in G$. Then $H=\langle a\rangle\leq G$. So, $a^m=e$ for some $m\in\mathbb{N}$.

Definition. If $G$ is finite, then the order of $a\ne e\in G$, written $|a|$ or $o(a)$, is the least positive integer $n$ such that $a^n=e$.

Corollary. If $G$ is finite and $a\ne e\in G$, then $|\langle a\rangle|||G|$.

Theorem. If $G$ is a finite group of order $n$, then $a^n=e$ for all $a\in G$.

Proof. Let $a\ne e\in G$ and $|\langle a\rangle|=k$. Then by Lagrange’s Theorem $k|n$ i.e. $n=kl$ for some $l\in\mathbb{N}$. So,
$$a^n=a^{kl}=(a^k)^l=e^l=e.$$

In this lecture, we study subgroups. A nonempty subset $H$ of a group $G$ is called a subgroup of $G$ if $H$ itself forms a group relative to the operation $\cdot$ in $G$. If $H$ is a subgroup of $G$, we simply write $H\leq G$. Clearly $G\leq G$ and $\{e\}\leq G$. $G$ and $\{e\}$ are called trivial subgroups of $G$. Before we discuss subgroups further we need to go over some basic properties of a group as we need them.

Lemma. If $G$ is a group, then:

(a) Its identity element is unique.

(b) Each $a\in G$ has a unique inverse $a^{-1}\in G$.

(c) $\forall a\in G$, $(a^{-1})^{-1}=a$.

(d) $\forall a,b\in G$, $(ab)^{-1}=b^{-1}a^{-1}$.

The proof of this lemma is pretty much straightforward and is left to readers.

In order to show that a nonempty subset $H$ of a group $G$ is a subgroup we need to check:

1. $H$ is closed under $\cdot$, the operation in $G$. That is, $\forall a,b\in H$, $ab\in H$.

2. $e\in H$.

3. $\forall a\in H$, $a^{-1}\in H$.

Note that the associative law holds automatically. It turns out that there is a simpler criterion to check if a nonempty subset is a subgroup of a group.

Lemma. A nonempty subset $H\subset G$ is a subgroup of $G$ if and only if $\forall a,b\in H$, $ab^{-1}\in H$.

Proof. ($\Longrightarrow$) Clear.

($\Longleftarrow$) Let $a,b\in H$. Then by assumption, $aa^{-1}=e\in H$. Since $a,e\in H$, $ea^{-1}=a^{-1}\in H$. Since $b^{-1}\in H$, $a(b^{-1})^{-1}=ab\in H$. Therefore, $H$ is a subgroup of $G$.

Example. $\forall n\in\mathbb{N}\cup\{0\}$, $n\mathbb{Z}\leq(\mathbb{Z},+)$. Here, $n\mathbb{Z}=\{nx: x\in\mathbb{Z}\}$. In fact, they are all the subgroups of $(\mathbb{Z},+)$.

Definition. The cyclic subgroup of $G$ generated by $a\in G$ is $\{a^k: k\in\mathbb{Z}\}$. It is denoted by $\langle a\rangle$.

Example. Let $G$ be a group and $a\in G$. Let $C(a)=\{g\in G: ga=ag\}$. Then $C(a)\leq G$. $C(a)$ is called the centralizer of $a$ in $G$.

Example. Let $G$ be a group and let $Z(G)=\{z\in G: zx=xz\ \forall x\in G\}$. Then $Z(G)$ is a subgroup of $G$, called the center of $G$.

Example. Let $G$ be a group and $H\leq G$. $\forall a\in G$, $a^{-1}Ha=\{a^{-1}ha: h\in H\}$ is also a subgroup of $G$.

Lemma. Suppose that $(G,\cdot)$ is a group and $H$ is a finite nonempty subset of $H$. If $H$ is closed under $\cdot$, then $H$ is a subgroup of $G$.

Proof. Let $H=\{a_1,a_2,\cdots,a_n\}$. Then $\forall a\in H$, $aH=\{aa_1,aa_2,\cdots,aa_n\}\subset H$. On the other hand, if $aa_i=aa_j$ then $a_i=a_j$. So, $|aH|=|H|$. This implies that $aH=H$ and hence
\begin{align*}
a\in aH &\Longrightarrow a=aa_i\ \mbox{for some}\ i\\
&\Longrightarrow a_i=e\in H.
\end{align*}
Since $e\in aH$, $e=aa_j$ for some $j$ $\Longrightarrow$ $a_j=a^{-1}$.

Example. The symmetric group $S_3=\{1,(123),(132),(23),(13),(12)\}$ has 5 subgroups $\{1\}$, $\{1,(12)\}$, $\{1,(13)\}$, $\{1,(23)\}$, and $A_3=\{1,(123),(132)\}$. $A_3$ is called alternating group of degree 3. Cycles of length 2 such as (12), (13), (23) are called transpositions. Any permutation can be written as a product of either an even number of transpositions or an odd number of transpositions. If a permutation is a product of an even number of transpositions, the permutation is called an even permutation. If a permutation is a product of an odd number of transpositions, the permutation is called an odd permutation. The identity permutation 1 is an even permutation as $\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3 \end{pmatrix}$ can be written as (12)(21)(23)(32)(31)(13). The permutation (123) can be written as (12)(13). So, (123) is an even permutation. Similarly (132) is an even permutation as well. So it turns out that the alternating group $A_3$ is the set of all even permutations. In general, the set $A_n$ of all even permutations of $\{1,2,3,\cdots,n\}$ form a subgroup of the symmetric group $S_n$. Remember that $S_n$ has order $n!$. Since there are exactly the same number of even permutations and odd permutations, the alternating group $A_n$ of degree $n$ has order $\frac{n!}{2}$.

Example. $\mathrm{SL}(n,\mathbb{R})=\{A\in\mathrm{GL}(n,\mathbb{R}): \det A=1\}$ is a subgroup of the general linear group $\mathrm{GL}(n,\mathbb{R})$ of degree $n$. $\mathrm{SL}(n,\mathbb{R})$ is called the special linear group of degree $n$.

In the first lecture here, we defined a group. We have also seen an example of a group here, which is a symmetric group. In this lecture, we study some well-known examples of groups of finite and infinite orders. Recall that the order of a group os the number of elements in the group.

Examples of Groups

1. $(\mathbb{Z},+)$, $(\mathbb{Q},+)$, $(\mathbb{R},+)$, and $(\mathbb{C},+)$ are abelian groups of infinite order.
2. $(\mathbb{Q}\setminus\{0\},\cdot)$ is an abelian group of infinite order.
3. $(\mathbb{R}^+,\cdot)$ is an abelian group of infinite order. Here $\mathbb{R}^+$ denotes the set of all positive real numbers.
4. Let $E_n=\{e^{\frac{2k\pi i}{n}}: k=0,1,2,\cdots,n-1\}$. $e^{\frac{2k\pi i}{n}}$, $k=0,1,2,\cdots,n-1$ are the $n$-th roots of unity i.e. the zeros of $z^n=1$. $E_n$ forms an abelian group of order $n$. It is generated by a single element $e^{\frac{2\pi i}{n}}$. Such a group is called a cyclic group.

Definition. A group $G$ is said to be a finite group if it has a finite number of elements. The number of elements in $G$ is called the order of $G$ as mentioned previously and it is denoted by $|G|$ or $\mathrm{ord}(G)$. We use the notation $|G|$ for the order of the group $G$.

Example.Let $\mathbb{Z}_n=\{0,1,2,\cdots,n-1\}$. $\mathbb{Z}_n$ is the set of all possible remainders when an integer is divided by 2. The addition $+$ on $\mathbb{Z}_n$ is defined naturally as follows.
$$\begin{aligned}
\mathbb{Z}_2\\
\begin{array}{|c||c|c|}
\hline
+ & 0 & 1\\
\hline
\hline
0 & 0 & 1\\
\hline
1 & 1 & 0\\
\hline
\end{array}
\end{aligned}$$
$$\begin{aligned}
\mathbb{Z}_3\\
\begin{array}{|c||c|c|c|}
\hline
+ & 0 & 1 & 2\\
\hline
\hline
0 & 0 & 1 & 2\\
\hline
1 & 1 & 2 & 0\\
\hline
2 & 0 & 1 & 2\\
\hline
\end{array}
\end{aligned}$$
$(\mathbb{Z}_n,+)$ is an abelian group of order $n$. We will see later that the group $E_n$ and $\mathbb{Z}_n$ are indeed the same group.

Example. The set $M_{m\times n}(\mathbb{R})$ of all $m\times n$ matrices of real entries under matrix addition is an abelian group of infinite order.

Example. The set $\mathrm{GL}(n,\mathbb{R})$ of alll $n\times n$ non-singular matrices of real entries under matrix multiplication is a non-abelian group of infinite order. $\mathrm{GL}(n,\mathbb{R})$ is called the general linear group of degree $n$. It can be viewed as the set of all invertible linear transformations $T: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ i.e. linear isomorphisms from $\mathbb{R}^n$ onto itself.

Example. [Klein four-group (Vierergruppe in German)]

Klein four-group

Let $V=\{e,a,b,c\}$ where $a$ is counterclockwise rotation of the rectangle in the picture about the $x$-axis by $180^\circ$, $b$ is counterclockwise rotation of the rectangle about the $y$-axis by $180^\circ$ and $c$ is counterclockwise rotation of the rectangle about the $z$-axis (the axis coming out of the origin toward you) by $180^\circ$. Then $a$, $b$ and $c$ satisfy the following relationship
\begin{align*}
a^2=b^2=c^2=e,\ ab&=ba=c,\ bc=cb=a,\\
ca&=ac=b.
\end{align*}
Here, $\cdot$ is the function composition i.e. successive application of rotations. By labling the vertices of the rectangle as 1, 2, 3, 4 as seen in the picture, we can define each rotation as a permutation of $\{1,2,3,4\}$.
\begin{align*}
e&=\begin{pmatrix}
1 & 2 & 3 & 4\\
1 & 2 & 3 & 4
\end{pmatrix},\ a=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 &3
\end{pmatrix},\\
b&=\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix},\ c=\begin{pmatrix}
1 & 2 & 3 & 4\\
3 & 4 & 1 &2
\end{pmatrix}.
\end{align*}
For instance, we calculate $ab$:
\begin{align*}
ab&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 &3
\end{pmatrix}\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 &3
\end{pmatrix}\begin{pmatrix}
4 & 3 & 2 & 1\\
3 & 4 & 1 & 2
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 2 & 3 & 4\\
3 & 4 & 1 &2
\end{pmatrix}\\
&=c.
\end{align*}

Example. [The $n$-th dihedral group $D_n$, $n\geq 3$]

Consider an equilateral triangle shown in the following picture.

Dihedral group D3

$\rho$ is counterclockwise rotation about the axis coming out of the origin by $\frac{360^\circ}{3}=120^\circ$ and $\mu_i$, $i=1,2,3$ is counterclockwise rotation about the axis of rotation through each vertex $i$ by $180^\circ$. As permutations of the vertices $\{1,2,3\}$, $\rho$ and $\mu_i$ , $i=1,2,3$ are given by
\begin{align*}
\rho&=\begin{pmatrix}
1 & 2 & 3\\
2 & 3 & 1
\end{pmatrix},\ \mu_1=\begin{pmatrix}
1 & 2 & 3\\
1 & 3 & 2
\end{pmatrix}\\
\mu_2&=\begin{pmatrix}
1 & 2 & 3\\
3 & 2 & 1
\end{pmatrix},\ \mu_3=\begin{pmatrix}
1 & 2 & 3\\
2 & 1 &3
\end{pmatrix}.
\end{align*}
$D_3=\{e,\rho,\rho^2,\mu_1,\mu_2,\mu_3\}$ is a group called the 3rd dihedral group. $$\mu_1\mu_2=\rho^2\ne\rho=\mu_2\mu_1.$$
So, we see that $D_3$ is not an abelian group. $D_3$ is the group of symmetries of an equilateral triangle.

Now this time let us consider a square as shown in the following picture.

Dihedral group D4

$\rho$ is counterclockwise rotation about the $z$-axis coming out of the origin by $\frac{360^\circ}{4}=90^\circ$. $\mu_i$, $i=1,2$ are counterclockwise rotations about $y$-axis and $x$-axis, respectively by $180^\circ$. $\delta_i$, $i=1,2$ are counterclockwise rotations about the axis through the vertices 2 and 4 and the vertices through 1 and 3, respectively by $180^\circ$. As permutations of the vertices $\{1,2,3,4\}$, $\rho$, $\mu_i$ and $\delta_i$, $i=1,2$ are given by
\begin{align*}
\rho&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 4 & 1
\end{pmatrix},\\
\mu_1&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 & 3
\end{pmatrix},\ \mu_2=\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix},\\
\delta_1&=\begin{pmatrix}
1 & 2 & 3 & 4\\
3 & 2 & 1 & 4
\end{pmatrix},\ \delta_2=\begin{pmatrix}
1 & 2 & 3 & 4\\
1 & 4 & 3 & 2
\end{pmatrix}.
\end{align*}
$D_4=\{e,\rho,\rho^2,\rho^3,\mu_1,\mu_2,\delta_1,\delta_2\}$ is the group of symmetries of a square, called the 4th dihedral group. It is also called the octic group. Note that $|D_n|=2n$, $n\geq 3$.