The group $\mathbb{Z}\times\mathbb{Z}_2$ is generated by $(1,0)$ and $(0,1)$. In general, the direct product of $n$ cyclic groups, each of which is either $\mathbb{Z}$ or $\mathbb{Z}_m$ is generated by $(1,0,\cdots,0)$, $(0,1,0,\cdots,0)$, $\cdots$, $(0,0,\cdots,0,1)$. Such a direct product may be generated by fewer elements. For example, $\mathbb{Z}_3\times\mathbb{Z}_4\times\mathbb{Z}_{35}$ is generated by the single element $(1,1,1)$ i.e. it is a cyclic group. Conversely, we have the following theorem holds.

Theorem. [Fundamental Theorem of Finitely Generated Abelian Groups] Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form
$$\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\cdots\times\mathbb{Z}_{p_n^{r_n}}\times\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z},$$where the $p_i$ are primes, not necessarily distinct and the $r_i$ are positive integers. The direct product is unique except for a possible rearrangement of the factors. The number of factors $\mathbb{Z}$ is called the Betti number of $G$.

Example. Find all abelian groups, up to isomorphism, of order 360.

Solution. $360=2^3\cdot 3^2\cdot 5$, so all abelian groups, up to isomorphism, of order 360 are
\begin{align*}
\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5,\\
\mathbb{Z}_2\times\mathbb{Z}_4\times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5,\\
\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_9\times\mathbb{Z}_5,\\
\mathbb{Z}_2\times\mathbb{Z}_4\times\mathbb{Z}_9\times\mathbb{Z}_5,\\
\mathbb{Z}_8\times\mathbb{Z}_3\times\mathbb{Z}_3\times\times\mathbb{Z}_5,\\
\mathbb{Z}_8\times\mathbb{Z}_9\times\mathbb{Z}_5.
\end{align*}

Definition. A group $G$ is decomposable if it is isomorphic to a direct product of two proper nontrivial subgroups. Otherwise $G$ is indecomposable.

Theorem. The finite indecomposable abelian groups are exactly the cyclic groups with order a power of a prime.

Proof. If $G$ is a finite indecomposable abelian group, then by Fundamental Theorem of Finitely Generated Abelian Groups, $G$ is isomorphic to a direct product of cyclic groups of prime power order.Since $G$ is indecomposable, the direct product must consist of just one cyclic group of a prime power order. conversely, let $p$ be a prime. Then $\mathbb{Z}_{p^r}$ is indecomposable. If it were isomorphic to $\mathbb{Z}_{p^i}\times\mathbb{Z}_{p^j}$ where $i+j=r$, then every element has an order at most $p^{\max(i,j)}<p^r$.

As noted here, the converse of Lagrange’s theorem need not be true in general, however it is true for abelian groups.

Theorem. If $m$ divides the order of a finite abelian group $G$, then $G$ has a subgroup of order $m$.

Proof. Since $G$ is a finite abelian group,
$$G\cong\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\cdots\times\mathbb{Z}_{p_n^{r_n}}.$$Since $p_1^{r_1}p_2^{r_2}\cdots p_n^{r_n}=|G|$, $m$ must be of the form $p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$, where $0\leq s_i\leq r_i$. For each $1\leq i\leq n$, $p_i^{r_i-s_i}$ generates a cyclic group of order$$\frac{p_i^{r_i}}{(p_i^{r_i},p_i^{r_i-s_i})}=\frac{p_i^{r_i}}{p_i^{r_i-s_i}}=p_i^{s_i}.$$So, $p_i^{r_i-s_i}$ generates a cyclic subgroup of $\mathbb{Z}_{p_i^{r_i}}$ of order $p_i^{s_i}$. Therefore,$$\langle p_1^{r_1-s_1}\rangle\times\langle p_2^{r_2-s_2}\rangle\times\cdots\times\langle p_n^{r_n-s_n}\rangle$$
is a subgroup of $G$ of order $m=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$.

It follows immediately from the above theorem that:

Corollary. [Sylow’s Theorem for Abelian Groups] Let $G$ be a finite abelian group. If $p$ is a prime number and $p^\alpha||G|$, then $G$ has a subgroup of order $p^\alpha$.

While it is more difficult to prove, the statement in the corollary is indeed true for any finite group $G$ and it is referred to as Sylow’s Theorem.

Theorem. [Sylow’s Theorem] Let $G$ be a finite group. If $p$ is a prime number and $p^\alpha||G|$, then $G$ has a subgroup of order $p^\alpha$.

Theorem. If $m$ is a square free integer i.e. if $m$ is not divisible by the square of any prime, then every abelian group of order $m$ is cyclic.

Proof. Let $G$ be an abelian group of square free order $m$. Then
$$G\cong\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\cdots\times\mathbb{Z}_{p_n^{r_n}},$$
where $m=p_1^{r_1}p_2^{r_2}\cdots p_n^{r_n}$. Since $m$ is square free, $r_i=1$, $i=1,\cdots,n$ and the $p_i$ are distinct. Hence, $G$ is isomorphic to $\mathbb{Z}_{p_1p_2\cdots p_n}$ i.e. it is cyclic.

Let $G_1,G_2,\cdots,G_n$ be groups. Consider the Cartesian product
$$\prod_{i=1}^nG_i:=G_1\times G_2\times\cdots\times G_n.$$
Define a binary operation $\cdot$ on $\prod_{i=1}^nG_i$ by
$$(a_1,a_2,\cdots,a_n)\cdot(b_1,b_2,\cdots,b_n)=(a_1b_1,a_2b_2,\cdots,a_nb_n)$$
for $(a_1,a_2,\cdots,a_n),(b_1,b_2,\cdots,b_n)\in\prod_{i=1}^nG_i$. Then $\cdot$ is well-defined. Clearly $\cdot$ is associative. $(e_1,e_2,\cdots,e_n)\in\prod_{i=1}^nG_i$ is an identity element. For each $(a_1,a_2,\cdots,a_n)\in\prod_{i=1}^nG_i$, $(a_1,a_2,\cdots,a_n)^{-1}=(a_1^{-1},a_2^{-1},\cdots,a_n^{-1})\in\prod_{i=1}^nG_i$. So, $\left(\prod_{i=1}^nG_i,\cdot\right)$ is a group called the direct product of the $G_i$’s. If the operation on each $G_i$ is commutative, refer to $\prod_{i=1}^nG_i$ as the direct sum of the groups $G_i$. In this case, we often use the notation $\bigoplus_{i=1}^nG_i$ instead of $\prod_{i=1}^nG_i$.

Example. Let $\mathbb{R}\oplus\mathbb{R}$ be the direct sum of $(\mathbb{R},+)$ and itself. Define a map $\varphi:\mathbb{R}\oplus\mathbb{R}\longrightarrow S^1\times S^1$ by
$$\varphi(x,y)=(e^{2\pi ix},e^{2\pi iy})$$
for each $(x,y)\in\mathbb{R}\oplus\mathbb{R}$. Then $\varphi$ is an onto-homomorphism. The kernel of $\varphi$ is
$$\ker\varphi=\mathbb{Z}\oplus\mathbb{Z}.$$
Hence, by the Fundamental Homomorphism Theorem
$$\mathbb{R}\oplus\mathbb{R}/\mathbb{Z}\oplus\mathbb{Z}\cong S^1\times S^1.$$That is, the quotient group $\mathbb{R}\oplus\mathbb{R}/\mathbb{Z}\oplus\mathbb{Z}$ is a torus. $\mathbb{R}\oplus\mathbb{R}/\mathbb{Z}\oplus\mathbb{Z}$ can be viwed as a quotient set $\mathbb{R}\oplus\mathbb{R}/\sim$ where $\sim$ is an equivalence relation on $\mathbb{R}\oplus\mathbb{R}$ defined as follows: For all $(x,y),(z,w)\in\mathbb{R}\oplus\mathbb{R}$,$$(x,y)\sim (z,w)\ \mbox{if}\ (x,y)-(z,w)=(m,n)$$for some $(m,n)\in\mathbb{Z}\times\mathbb{Z}$.

The following theorem is introduced without a proof.

Theorem. Let $(a_1,\cdots,a_n)\in\prod_{i=1}^nG_i$. If for each $i=1,\cdots,n$, $a_i$ is of finite order $r_i$ in $G_i$, then the order of $(a_1,\cdots,a_n)$ in $\prod_{i=1}^nG_i$ is the least common multiple of $r_1,r_2,\cdots,r_n$.

Example. Find the order of $(8,4,10)$ in $\mathbb{Z}_{12}\times\mathbb{Z}_{60}\times\mathbb{Z}_{24}$.

Solution. First we find the orders of 8, 4, 10 in $\mathbb{Z}_{12}$, $\mathbb{Z}_{60}$, and $\mathbb{Z}_{24}$, respectively. For that let us recall a theorem we studied here. The theorem can be restated for an additive group as:

Theorem. Let $G$ be a finite additive group and $a\in G$ with $|a|=n$. Then for any $k\in\mathbb{Z}$,

1.  $|ka|=\frac{|a|}{(k,|a|)}$.
2.  $|ka|=n$ if and only if $(k,|a|)=1$.

Since 1 has order $n$ in $\mathbb{Z}_n$, we have the following corollary.

Corollary. The order of $1\leq k\leq n-1$ in $\mathbb{Z}_n$ is $\frac{n}{(k,n)}$.

It follows from this corollary that the order of 8 in $\mathbb{Z}_{12}$ is $\frac{12}{(8,12)}=\frac{12}{4}=3$, the order of 4 in $\mathbb{Z}_{60}$ is $\frac{60}{(4,60)}=\frac{60}{4}=15$, and the order of 10 in $\mathbb{Z}_{24}$ is $\frac{24}{(10,24)}=\frac{24}{2}=12$. The least common multiple of 3, 15, 12 is 60, so the order of $(8,4,10)$ in $\mathbb{Z}_{12}\times\mathbb{Z}_{60}\times\mathbb{Z}_{24}$ is 60.

Example. $\mathbb{Z}_2\times\mathbb{Z}_3=\{(0,0), (0,1), (0,2),(1,0),(1,1),(1,3)\}$ is a cyclic group generated by $(1,1)$. Hence, $\mathbb{Z}_2\times\mathbb{Z}_3\cong\mathbb{Z}_6$.

Example. $\mathbb{Z}_2\times\mathbb{Z}_2$ is not cyclic. $\mathbb{Z}_2\times\mathbb{Z}_2\cong V_4$, Klein four-group.

Theorem. $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic and isomorphic to $\mathbb{Z}_{mn}$ if and only if $(m,n)=1$.

Corollary. $\prod_{i=1}^n\mathbb{Z}_{m_i}$ is cyclic and isomorphic to $\mathbb{Z}_{m_1m_2\cdots m_n}$ if and only if the numbers $m_i$ for $i=1,2,\cdots,n$ are such that the greatest common divisor of any two of them is 1.

Corollary. If $n=(p_1)^{n_1}(p_2)^{n_2}\cdots(p_r)^{n_r}$ where $p_1,p_2,\cdots,p_r$ are distinct primes, then
$$\mathbb{Z}_n\cong\mathbb{Z}_{(p_1)^{n_1}}\times\mathbb{Z}_{(p_2)^{n_2}}\times\cdots\times\mathbb{Z}_{(p_r)^{n_r}}.$$

Example. $\mathbb{Z}_8\times\mathbb{Z}_9\cong\mathbb{Z}_{72}$.

Let $\prod_{i=1}^nG_i$ be the direct product of groups $G_1,\cdots,G_n$. For each $i=1,\cdots,n$, let
$$\bar G_i=\{(e_1,e_2,\cdots,e_{i-1},a_i,e_{i+1},\dots,e_n): a_i\in G_i\}\leq\prod_{i=1}^n G_i.$$
Then for each $i=1,\cdots,n$, $\bar G_i\cong G_i$. The direct product $\prod_{i=1}^n\bar G_i$ of the groups $\bar G_1,\cdots,\bar G_n$ is called an internal direct product while $\prod_{i=1}^nG_i$ is called an external direct product. Clearly the external and internal direct products are isomorphic to each other.

The following theorem is called the Fundamental Homomorphism Theorem or the First Isomorphism Theorem.

Theorem. Let $G$ and $G’$ be groups, and $\varphi:G\longrightarrow G’$ an epimorphism (onto homomorphism). Then $G/K\cong G’$ where $K=\ker\varphi$.

Proof.

Let $\gamma: G\longrightarrow G/K$ be the canonical homomorphism i.e. for any $a\in G$, $\gamma(a)=Ka$. Define $\psi: G/K\longrightarrow G’$ by
$$\psi(Ka)=\varphi(a)$$for each $a\in G$. Then

1. $\psi$ is well-defined:\begin{align*}Ka=Kb&\Rightarrow ab^{-1}\in K\\&\Rightarrow\varphi(ab^{-1})=e’\\&\Rightarrow\psi(Ka)=\varphi(a)=\varphi(b)=\psi(Kb).\end{align*}
2. $\psi$ is a homomorphism:$$\psi(KaKb)=\psi(Kab)=\varphi(ab)=\varphi(a)\varphi(b)=\psi(Ka)\psi(Kb).$$
3. $\psi$ is one-to-one:\begin{align*}\psi(Ka)=\psi(Kb)&\Rightarrow\varphi(a)=\varphi(b)\\&\Rightarrow\varphi(ab^{-1})=e’\\&\Rightarrow ab^{-1}\in K\\&\Rightarrow Ka=Kb.\end{align*}
4. $\psi$ is onto: Let $b\in G’$. Then there exists $a\in G$ such that $\varphi(a)=b$ since $\varphi$ is onto. Now, $Ka\in G/K$ and $\psi(Ka)=\varphi(a)=b$.

Example. Let $S^1$ be the unit circle centered at the origin. Then it can be represented in terms of complex numbers as
$$S^1=\{e^{2x\pi i}:x\in[0,1)\}.$$Define a map $\varphi: (\mathbb{R},+)\longrightarrow(S^1,\cdot)$ by $$\varphi(x)=e^{2\pi ix}$$for each $x\in\mathbb{R}$. Then $\varphi$ is an onto homomorphism. The kernel of $\varphi$ is$$\ker\varphi=\mathbb{Z}.$$Hence, by the Fundamental Homomorphism Theorem
$$\mathbb{R}/\mathbb{Z}\cong S^1.$$Note that $\mathbb{R}/\mathbb{Z}$ can be viewed as the quotient set $\mathbb{R}/\sim$ where $\sim$ is an equivalence relation on $\mathbb{R}$ defined as follows: For all $x,y\in\mathbb{R}$,
$$x\sim y\ \mbox{if}\ x-y=n$$for some $n\in\mathbb{Z}$.

Theorem [Correspondence Theorem]. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Let $K=\ker\varphi$, $H’\leq G’$ and $H=\varphi^{-1}(H’)=\{a\in G: \varphi(a)\in H\}$. Then $K\subset H\leq G$ and $H/K\cong H’$. If $H’\triangleleft G’$, then $H\triangleleft G’$.

Proof. Since $e\in K\subset H$, $H\ne\emptyset$. Let $a,b\in H$. Then $\varphi(a),\varphi(b)\in H’$. Since $H’\leq G’$, $\varphi(a)\varphi(b)^{-1}=\varphi(ab^{-1})\in H’$ and so, $ab^{-1}\in H$. Hence, $H\leq G$. Since $e’\in H’$, $K\subset H$. Let $\psi=\varphi|_H$. Then $\psi$ is a homomorphism from $H$ onto $H’$ and $\ker\psi=\ker\varphi=K$. Therefore, by the Fundamental Homomorphism Theorem $H/K\cong H’$.

Suppose that $H’\triangleleft G’$. Let $a\in G$ and $h\in H$. Then $\varphi(a)\in G’$ and $\varphi(h)\in H’$. Since $H’\triangleleft G’$, $\varphi(a)\varphi(h)\varphi(a)^{-1}=\varphi(aha^{-1})\in H’$ which implies that $aha^{-1}\in H$. Hence, $H\triangleleft G$.

Theorem [The Second Isomorphism Theorem]. Let $H\leq G$ and $N\triangleleft G$. Then $HN\leq G$, $H\cap N\triangleleft H$ and
$$H/H\cap N\cong HN/N.$$

Proof. Let $a,b\in HN$. Then $a=h_1n_1$ and $b=h_2n_2$ for some $h_1,h_2\in H$ and $n_1,n_2\in N$. Then
\begin{align*}
ab^{-1}&=h_1n_1(h_2n_2)^{-1}\\
&=h_1n_1(n_2^{-1}h_2^{-1})\\
&=h_1h_2^{-1}(h_2(n_1n_2^{-1})h_2^{-1})\in HN.
\end{align*}
So, $HN\leq G$. Clearly $H\cap N\triangleleft H$. Also clearly $N\leq HN$. Let $n_1\in N$. Then $\forall hn\in HN$,
$$(hn)n_1(hn)^{-1}=h(nn_1n^{-1})h^{-1}\in N.$$This means that $N\triangleleft HN$. Define $\varphi:H\longrightarrow HN/N$ by
$$\varphi(h)=Nh$$for each $h\in H$. Then $\varphi$ is clearly well-defined, a homomorphism. Let $Nhn\in HN/N$. Then $Nhn=hnN=hN=Nh$ and $\varphi(h)=Nh=Nhn$. So, $\varphi$ is onto.
\begin{align*}
h\in \ker\varphi&\Leftrightarrow \varphi(h)=Nh=N\\
&\Leftrightarrow h\in N\\
&\Leftrightarrow h\in H\cap N.
\end{align*}
So, $\ker\varphi=H\cap N$. Therefore, by the Fundamental Homomorphism Theorem
$$H/H\cap N\cong HN/N.$$

Theorem [The Third Isomorphism Theorem]. Let $\varphi: G\longrightarrow G’$ be an epimorphism. Let $K=\ker\varphi$, $N’\triangleleft G’$, and $N=\varphi^{-1}(N’)=\{a\in G: \varphi(a)\in N’\}$. Then
$$G/N\cong G’/N,$$or equivalently
$$G/N\cong(G/K)/(N/K).$$

Proof. Define $\psi: G\longrightarrow G’/N’$ by
$$\psi(a)=N’\varphi(a)$$for each $a\in G$. Then

1. $\psi$ is well-defined:\begin{align*}a=b\in G &\Rightarrow\varphi(a)=\varphi(b)\\&\Rightarrow \psi(a)=N’\varphi(a)=N’\varphi(b)=\psi(b).\end{align*}
2. $\psi$ is a homomorphism:\begin{align*}\psi(ab)&=N’\varphi(ab)\\&=N’\varphi(a)\varphi(b)\\&=N’\varphi(a)N’\varphi(b)\\&=\psi(a)\psi(b).\end{align*}
3. $\psi$ is onto: Let $N’c\in G’/N’$. Then $c\in G’$. Since $\varphi$ is onto, there exists $a\in G$ such that $\varphi(a)=c$. $N’c=N’\varphi(a)=\psi(a)$.
4. $\ker\psi=N$:\begin{align*}a\in\ker\psi&\Leftrightarrow N’\varphi(a)=N’\\&\Leftrightarrow\varphi(a)\in N’\\&\Leftrightarrow a\in N.\end{align*}

Therefore, by the Fundamental Homomorphism Theorem we obtain
$$G/N\cong G’/N’.$$Since $\varphi:G\longrightarrow G’$ is an epimorphism, by the Fundamental Homomorphism Theorem, $G’\cong G/K$. Since $N=\varphi^{-1}(N’)$, by the Correspondence Theorem, $N’\cong N/K$. Hence, $G’/N’\cong(G/K)/(N/K)$ i.e. $G/N\cong(G/K)/(N/K)$.

Let $G$ be a group and $N\leq G$. Recall the equivalence relation $\sim$ on $G$ defined by
$$a\sim b\ \mbox{if}\ ab^{-1}\in N$$ for any $a,b\in G$. Each equivalence class $[a]$ is identified with right coset $Na$. The quotient set $G/\sim$ is the set of all equivalence classes or equivalently all right cosets of the subgroup $N$ in $G$. A question we can ask is can we give a group structure to the quotient set $G/\sim$? The answer is affirmative if $N$ is a normal subgroup of $G$. This is what we are going to discuss in this lecture.

Let $N\vartriangleleft G$. Denote by $G/N$ the set of all right cosets of $N$ in $G$ i.e.
$$G/N=\{Na:a\in G\}.$$
For any $Na, Nb\in G/N$,
\begin{align*}
(Na)(Nb)&=N(aN)b\\
&=N(Na)b\\
&=N(Nab)\\
&=Nab.
\end{align*}
So, it appears that if $N\vartriangleleft G$, we may define an operation $\cdot$ on $G/N$ naturally from the binary operation on $G$ by the equation \begin{equation}
\label{eq:cosetoper}
Na\cdot Nb=Nab
\end{equation} for any $Na,Nb\in G/N$. But is this operation well-defined? To see that let $Na=Nc$ and $Nb=Nd$. Then $Nac^{-1}=N$ and $Nbd^{-1}=N$. So $ac^{-1}\in N$ and $bd^{-1}\in N$. Now,
\begin{align*}
N(ac)(bd)^{-1}&=N(ac)(d^{-1}b^{-1})\\
&=Na(cd^{-1})b^{-1}\\
&=a(Ncd^{-1})b^{-1}\\
&=aNb^{-1}\\
&=Nab^{-1}\\
&=N.
\end{align*}
That is, $Na\cdot Nb=Nab=Ncd=Nc\cdot Nd$. Hence, right coset operation given by the equation \eqref{eq:cosetoper} is well-defined. Conversely, if $N\leq G$ and right coset operation given by the equation \eqref{eq:cosetoper} is well-defined, then $N$ must be a normal subgroup of $G$. The verification of this is left to readers as an exercise. Furthermore, $Ne=N$ is an identity element in $G/N$ and for each $Na\in G/N$, $Na^{-1}\in G$ and $NaNa^{-1}=N$, hence $(Na)^{-1}=Na^{-1}$. Therefore, $(G/N,\cdot)$ is a group. This group is called a factor group or a quotient group of $G$ modulo $N$.

Theorem. If $N\vartriangleleft G$ then there exists an onto homomorphism (an epimorphism) $\psi:G\longrightarrow G/N$ such that $\ker\psi=N$, Such a homomorphism is called a natural homomorphism or a canonical homomorphism.

Proof. Define $\psi: G\longrightarrow G/N$ by
$$\psi(a)=Na$$
for any $a\in G$. If $a=b$ then $\psi(a)=Na=Nb=\psi(b)$ so $\psi$ is well-defined. Let $Na\in G/N$. Then $a\in G$ and $\psi(a)=Na$, so $\psi$ is onto. $\psi(ab)=Nab=NaNb=\psi(a)\psi(b)$, hence $\psi$ is a homomorphism. To show that $\ker\psi=N$,
\begin{align*}
a\in\ker\psi&\Longleftrightarrow \psi(a)=N\\
&\Longleftrightarrow Na=N\\
&\Longleftrightarrow a\in N.
\end{align*}

$G/N$ is the set of all right coset of $N$ in $G$, so $|G/N|=|G:N|=\frac{|G|}{|N|}$.

Example. Since $\mathbb{Z}$ is an abelian group, for any $n\in\mathbb{N}$, $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$. So, for any $n\in\mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ is a factor group. Each $\mathbb{Z}/n\mathbb{Z}$ is indeed isomorphic to $\mathbb{Z}_n$.

The notion of factor groups can be used to prove an important theorem in group theory called Cauchy Theorem. First we study the following theorem as we will also need it (actually its corollary) to prove Cauchy Theorem.

Theorem. Let $G$ be a finite group and $a\in G$ with $|a|=n$. Then for any $k\in\mathbb{Z}$,

1.  $|a^k|=\frac{n}{d}$ where $d=(k,n)$.
2.  $|a^k|=n$ if and only if $(k,n)=1$.

Proof. Since 2 follows from 1, we prove 1 only. Since $d=(k,n)$, $n=n_1d$ and $k=k_1d$ for some $n_1,k_1\in\mathbb{Z}$ such that $(n_1,k_1)=1$.
$$(a^k)^{n_1}=a^{k_1n}=(a^n)^{k_1}=e.$$ So, if we let $|a^k|=s$, then $s|n_1$. On the other hand, $e=(a^k)^s=a^{ks}$, so $n|ks$ which implies that $n_1|k_1s$. Since $(n_1,k_1)=1$, $n_1|s$. Therefore, $s=n_1=\frac{n}{d}$.

Corollary. If a finite group $G$ has no nontrivial subgroup, then $G$ must be a cyclic group of a prime order.

Proof. Let $a\ne e\in G$. Then $\langle a\rangle\leq G$. Since $G$ has no nontrivial subgroup and $a\ne e$, $\langle a\rangle=G$ i.e. $G$ is a cyclic group. Let $|a|=n$. If $k\ne n$ and $k|n$. Then by the above theorem $|a^k|=\frac{n}{k}$ since $(n,k)=k$. Since $a^k\ne e$ and $G$ has no nontrivial subgroup, $\langle a^k\rangle=G$. This implies that $|a^k|=n$ and so $k=1$. Therefore, $n$ is a prime.

Theorem. [Cauchy] If $G$ is a finite abelian group and and $p||G|$ where $p$ is a prime, then $G$ has an element of order $p$.

Proof. We prove the theorem by induction on $|G|$. If $|G|=1$, then there is no prime that divides $|G|$, so the theorem is vacuously true. Suppose that the statement is true for all abelian groups whose order is less than $|G|$. Let $\{e\}\lneqq N\lneqq G$. If $p||N|$, then by induction hypothesis there exists $a\in N$ such that $|a|=p$. Since $N\subset G$, we are done in this case. Now suppose that $p\not||N|$. Since $G$ is abelian, $N\vartriangleleft G$. Since $p||G|$ and $p\not| |N|$, $p||G/N|<|G|$. So by induction hypothesis, $G/N$ has an element $Na$ of order $p$.
$$(Na)^p=Na^p=N\Longrightarrow a^p\in N$$
but $a\not\in N$ since $Na\ne N$. Let $|N|=m$. Then $e=(a^p)^m=(a^m)^p$. Let $b=a^m$. If $b\ne e$, then $b$ is an element of order $p$. What if $b=a^m=e$? If so, $(Na)^m=N$. Since $|Na|=p$, $p|m=|N|$. but by assumption $p\not||N|$. A contradiction! So, we are done if $G$ has a nontrivial subgroup. What if $G$ does not have a nontrivial subgroup? If so, by corollary above $G$ must be a cyclic group of a prime order. Since $p||G|$, $|G|=p$. In this case, any $a\ne e\in G$ is an element of order $p$.

Remark. It should be noted that Cauchy Theorem still holds even if the group $G$ is not abelian. We will not however prove the generalized version of Cauchy Theorem here.

In here, it was shown that the kernel $K=\ker\varphi$ of a homomorphism $\varphi: G\longrightarrow G’$ is a subgroup of $G$ and that it satisfies the property $aK=Ka$ for all $a\in G$. The kernel of a homomorphism is an example of a particular kind of subgroups of a group called normal subgroups.

Definition. The subgroup $N$ of $G$ is called a normal subgroup of $G$ if $a^{-1}Na\subset N$, $\forall a\in G$. If $N$ is a normal subgroup of $G$, we write $N\vartriangleleft G$.

Note that $\forall a\in G$, $a^{-1}Na\subset N$ implies that $\forall a\in G$, $a^{-1}Na=N$. Hence, we have

Theorem. $N\vartriangleleft G$ if and only if $\forall a\in G$, $aN=Na$.

If $G$ is abelian, every subgroup is a normal subgroup. But the converse need not be true. There are nonabelian groups in which every subgroup is normal. Such nonabelian groups are called Hamiltonian.

Example. Let $\mathbb{Q}_8=\{\pm 1,\pm, i, \pm j, \pm k\}$ where $1,i,j,k$ satisfy the multiplication laws
\begin{align*}
1^2&=1,\ i^2=j^2=k^2=-1,\\
ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j.
\end{align*}
Then $\mathbb{Q}_8$ forms a nonabelian group. There are four subgroups of $\mathbb{Q}_8$:
\begin{align*}
\langle -1\rangle&=\{1,-1\},\\
\langle i\rangle&=\{1,-1,i,-i\},\\
\langle j\rangle&=\{1,-1,j,-j\},\\
\langle k\rangle&=\{1,-1,k,-k\}.
\end{align*}
All these subgroups are normal subgroups of $\mathbb{Q}_8$, so $\mathbb{Q}_8$ is Hamiltonian. Note that $1,i,j,k$ form a 4-dimensional algebra of quaternions over $\mathbb{R}$ $$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$ The real algebra of quaternions $\mathbb{H}$ is identified with the 4-dimensional Euclidean space $\mathbb{R}^4$.

Recall that the index $|G:H|$ of a subgroup $H$ in $G$ is the number of right cosets of $H$ in $G$ or equivalently the number of left cosets of $H$ in $G$ as discussed here.

Theorem. Let $G$ be a group and $N\leq G$. If $|G:N|=2$ then $N\vartriangleleft G$.

Proof. Let $a\in G$. If $a\in N$, $aN=Na$. Now suppose that $a\in G\setminus N$. Then
$$N\cap Na=\emptyset=N\cap aN.$$
Furthermore, since $G=N\cup Na=N\cup aN$, $Na=Na$.

Example. Recall that $|S_n|=n!$ and $|A_n|=\frac{n!}{2}$. Since $|S_n:A_n|=2$, $A_n\vartriangleleft S_n$.

Remark. The converse of the above theorem need not be true, namely the index $|G:N|$ of a normal subgroup $N$ in $G$ is not necessarily 2. The 4th dihedral group (we introduced it here)
$$D_4=\{1,\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau\}$$
has the following nontrivial subgroups:
\begin{align*}
&\{1,\sigma^2,\sigma\tau,\sigma^2\tau\},\ \{1,\sigma,\sigma^2,\sigma^3\},\ \{1,\sigma^2,\tau,\sigma^2\tau\},\\
&\{1,\sigma^3\tau\},\ \{1,\sigma\tau\},\ \{1,\sigma^2\},\ \{1,\tau\},\ \{1,\sigma^2\tau\}.
\end{align*}
The subgroups of order 4 are normal subgroups of $D_4$ because their indices are 2. The center of $D_4$ is $\{1,\sigma^2\}$, so it is also a normal subgroup of $D_4$. However, $|D_4:Z(D_4)|=4$. The rest of nontrivial subgroups are not normal subgroups of $D_4$.

Remark. $H\vartriangleleft N, N\vartriangleleft G\not\Longrightarrow H\vartriangleleft G$. In the previous remark, $H=\{1,\tau\}$ is a normal subgroup of $\{1,\sigma^2,\tau,\sigma^2\tau\}$ but $H\not\vartriangleleft D_4$.

Example. The center $Z(G)$ of a group $G$
$$Z(G)=\{x\in G: \forall g\in G, xg=gx\}$$ is a normal subgroup of $G$.

Example. If $N\leq Z(G)$ then $N\vartriangleleft G$.

Example. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\vartriangleleft G$ as seen here.

Example. Let $\mathbb{R}^\ast=\mathbb{R}\setminus\{0\}$. Define $f: \mathrm{GL}(2,\mathbb{R})\longrightarrow\mathbb{R}^\ast$ by
$$f(A)=\det A,\ \mbox{for any}\ A\in\mathrm{GL}(2,\mathbb{R}).$$ Then $f$ is a homomorphism because for any $A,B\in\mathrm{GL}(2,\mathbb{R})$, $f(AB)=\det(AB)=(\det A)(\det B)=f(A)f(B)$. $f$ is also onto: for any $r\in\mathbb{R}^\ast$ let $A=\begin{pmatrix}
r & 0\\
0 & 1 \end{pmatrix}$. Then $A\in\mathrm{GL}(2,\mathbb{R})$ and $f(A)=r$. For any $A\in\mathrm{GL}(2,\mathbb{R})$, $A\in\ker f$ if and only if $\det A=1$ and so $\ker f=\mathrm{SL}(2,\mathbb{R})$. Hence, $\mathrm{SL}(2,\mathbb{R})\vartriangleleft\mathrm{GL}(2,\mathbb{R})$.