We study the properties of a subgroup of a group. We mention that group theory is an important part of algebra that classifies sets. A group is a particular set with a particular structure governed by a law of composition which makes it possible to carry out calculations. In this article, we make this concept clear with definitions, properties, and great exercises. Groups inter in the study of vector spaces and rings

**What is a group in algebra?**

In this section, we give a concise background on group theory. The most important properties of groups will be discussed.

**Definition:** Let $G$ be a no-empty set. A composition law on $G$ is an application $\ast: G\times G\to G$ such that for any $(x,y)\in G$ we have $x\ast y\in G$. This means that the composition of two elements of $G$ is still an element of $G$. This is a somehow stability property of $\ast$.

**Definition:** A set $(G,\ast)$ is called a group if $\ast$ is a composition law with the following properties:

**associativity**of $\ast$: for any $x,y$ and $z$ in $G$, $x\ast (y\ast z)=(x\ast y)\ast z$.- there exists an element $e\in G$ such that $x\ast e=e\ast x=x$. This element $e$ is called the
**neutral**element, - For any $x\in G$ there exists $y\in G$ such that $x\ast y=y\ast x=e$. The element $y$ is called the
**inverse**of $x$ and will be denoted by $x^{-1}$.

A group $(G,\ast)$ is called **commutative** group if for any $x,y\in G$, $x\ast y=y\ast x$.

$(\mathbb{R},+)$, here $\ast=+$ the usual addition operation, is a commutative group. Note that $(\mathbb{N},+)$ is not a group, because any element of $\mathbb{N}$ has no inverse.

If $(G,\ast)$ and $(H,\star)$ are two groups, then we can define another group $(G\times H, \diamond )$ by introducing the following composition law \begin{align*} (x,y)\diamond (x’,y’)=(x\ast x’,y\star y’).\end{align*}

**What is a subgroup of a group**?

**Notation:** In what follow we will denote any composition law on any set by “$\cdot$” and we also denote $x\cdot y=xy$.

A set $H$ is called a **subgroup** of a group $(G,\cdot)$ if $H\subset G$ and $(H,\cdot)$ is a group. This is equivalent to

- $H\neq \emptyset$
- $x,y\in H$ implies that $xy\in H,$
- for any $x\in H,$ the inverse $x^{-1}\in H$.

The last two assertions can be combined into a single assertion of the form $xy^{-1}\in H$.

**Proposition:** The intersection of two subgroups is a subgroup.

**A selection of exercises for groups **

Let us now give a selection of exercises on group theory.

**Exercise:** Does the following groups are isomorphs?

- $\mathbb{Z}/6\mathbb{Z}$ and $\mathfrak{S}_3,$ the group of permutations.
- $\mathbb{Z}/(nm)\mathbb{Z}$ and $\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ for $n,m\in\mathbb{N}$.
- $(\mathbb{R},+)$ and $(\mathbb{Q},+)$.
- $(\mathbb{R},+)$ and $(\mathbb{R}^\ast_{+},\times)$.
- $(\mathbb{Q},+)$ and $(\mathbb{Q}^\ast_{+},\times)$.

**Solution:** 1) The groups $\mathbb{Z}/6\mathbb{Z}$ and $\mathfrak{S}_3$ are not isomorphic because one is commutative and not the other.

2) When $n$ and $m$ are not coprime “relatively prime”, the groups $\mathbb{Z}/(nm)\mathbb{Z}$ and $\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ are not isomorphs. In fact if $p=gcd(n,m)$, then we must have $0 < p < nm$. Hence if we denote $[x]_r$ the elements of $\mathbb{Z}/r\mathbb{Z}$, for $r\in \mathbb{N}$, we have $p[1]_n=[p]_n\neq [0]_{nm}$.

If we assume that $\mathbb{Z}/(nm)\mathbb{Z}$ and $\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ are isomorph, then there would be an element $(x,y)\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ such that $p(x,y)\neq ([0]_n,[0]_m)$, which is absurd since $px=[0]_n$ “$p$ is a multiple of $n$” and $py=[0]_m$ “$p$ is a multiple of $m$”.

3) $(\mathbb{R},+)$ and $(\mathbb{Q},+)$ are not isomorph. In fact, we that $\mathbb{R}$ is not countable while $\mathbb{Q}$ is.

4) It is well known that the application $f:(\mathbb{R},+)\to \mathbb{R}^\ast_+$ such that $f(x)=e^x$ is bijective and $f(x+y)=e^{x+y}=e^x e^y=f(x)f(y)$ for any $x,y\in\mathbb{R}$. This shows that $f$ is an isomorphism of groups. Hence the groups $(\mathbb{R},+)$ and $(\mathbb{R}^\ast,\times)$ are isomorph.

5) The group $(\mathbb{Q},+)$ satisfies the following property \begin{align*} \forall y\in \mathbb{Q},\qquad \exists,x\in \mathbb{Q},\quad \text{s.t.}\;y=x+x. \end{align*} Now if there exists an isomorphism of groups between $(\mathbb{Q},+)$ and $(\mathbb{Q}^\ast_+,\times)$ then we will have \begin{align*} \forall y\in \mathbb{Q}^\ast_+,\qquad \exists,x\in \mathbb{Q}^\ast_+,\quad \text{such that}\;y=x\times x. \end{align*} This is not possible as e.g. the number $2$ does not admit a square root in $\mathbb{Q}$.

**Exercise:** We denote by $GL_2(\mathbb{R})$ the group of invertible matrices. Let \begin{align*} \mathcal{H}=\left\{\begin{pmatrix} 3^n&n3^{n-1}\\ 0&3^n\end{pmatrix}:n\in \mathbb{Z}\right\} \end{align*} a subgroup of $GL_2(\mathbb{R})$. Prove that $mathcal{H}$ is isomorph to $\mathbb{Z}$.

**Solution:** We shall use the fact that any infinite monogenic group is isomorph to $\mathbb{Z}$. We select \begin{align*} A=\begin{pmatrix} 3&1\\0&3\end{pmatrix}\in GL_2(\mathbb{R}). \end{align*} By using a recurrence argument one can see that \begin{align*} A^n=\begin{pmatrix} 3^n&n3^{n-1}\\ 0&3^n\end{pmatrix},\quad n\in \mathbb{Z}. \end{align*} Then \begin{align*} \mathcal{H}=\{A^n:n\in\mathbb{Z}\}. \end{align*} Then $\mathcal{H}$ is an infinite monogenic group, so it is isomorph to $\mathbb{Z}$.

**Exercise:** Let $G$ be a group “supposed not Abelian”. To any $a\in G,$ we associate an application \begin{align*} f_a:G\to G,\quad x\mapsto f_a(x)=axa^{-1}. \end{align*}

- Prove that $f_a$ is an isomorphism from $G$ to $G$.
- We denote \begin{align*} {\rm Int}(G)=\{f_a:a\in G\}. \end{align*} Prove that $({\rm Int}(G),\circ)$ is a group.
- Prove that \begin{align*} \psi: G\to {\rm Int}(G),\quad a\mapsto f_a \end{align*} is an homomorphism of groups. Determine its kernel.

**Solution:** 1) We denote by $e$ the identity element of $G$, we then have $f_a(e)=aea^{-1}=aa^{-1}=e$. For $x,y\in G$, we have \begin{align*} f_a(xy)&=axya^{-1}=axeya^{-1}=axa^{-1}aya^{-1}\cr &=(axa^{-1})(aya^{-1})\cr &= f_a(x)f_a(y).\end{align*} Then $f_a$ is a homomorphism of groups. Let us prove that $f_a$ is bijective. It suffices to show that for any $y\in G$ there is a unique $x\in G$ such that $f_a(x)=y$, which means that $axa^{-1}=y$. This is equivalent to $x=a^{-1}ya$. This element is unique, so $f_a$ is an isomorphism.

2) We denote by $(B(G),\circ)$ the group of all isomorphism from $G$ to $G$. Observe that ${\rm Int}(G)\subset B(G)$. It suffices then to show that ${\rm Int}(G)$ is a subgroup of $B(G)$. In fact, remark that for any $x\in G$, we have ${\rm id}_G(x)=x=exe^{-1}=f_e(x)$. This means that ${\rm id}_G\in {\rm Int}(G)$, and then ${\rm Int}(G)$ is not empty. Let $\ell_1,\ell_2\in {\rm Int}(G)$. Then, there exist $a,b\in G$ such that $\ell_1=f_a$ and $\ell_2=f_b$. Now for any $x\in G,$ we have \begin{align*} (\ell_1\circ\ell_2)(x)&=\ell_1(\ell_2(x))\cr &=f_a(f_b(x))\cr &= af_b(x)a^{-1}\cr &= abxb^{-1}a^{-1}\cr &=(ab)x(ab)^{-1}\cr &= f_{ab}(x). \end{align*} As $ab\in G,$ then $f_{ab}\in {\rm Int}(G)$. Hence $\ell_1\circ\ell_2\in {\rm Int}(G)$. According to the proof of the first question, for any $a\in G,$ we have $f^{-1}_a=f_{a^{-1}}\in {\rm Int}(G),$ because $a^{-1}\in G$. This ends the proof.

3) In the proof of the second question we have seen that for $a,b\in G$ we gave $f_a\circ f_b=f_{ab}$. Hence \begin{align*} \psi(ab)&=f_{ab}=f_a\circ f_b\cr &= \psi(a)\circ \psi(b). \end{align*} This shows that $\psi$ is a homomorphism of groups.

Let $a\in \ker(\psi)$, which means that $f_a={\rm id}_G$. For all $x\in G,$ $f_a(x)=x$, then $axa^{-1}=x$. This implies that $a$ satisfies $ax=xa$ for all $x\in G$. Hence \begin{align*} \ker(\psi)=\{a\in G\;|\; ax=xa,\;\forall x\in G\}. \end{align*}