# Group theory: course and worksheet for beginners

Group theory is a big part of advanced algebra. The Group is a special set with a particular structure (the set of real numbers is a particular group). A set is called a group if it has a law of composition which allows calculations to be made. In this article, we make this concept clear with definitions, properties, and great exercises.

### A short course

In this section, we give a concise course 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 $x\ast y\in G$.

Definition: A set $(G,\ast)$ is called a group is $\ast$ 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 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.

If $(H,\ast)$ 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*}

### Group theory worksheet

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

Exercise: Does the following groups are isomorph?

• $\mathbb{Z}/6\mathbb{Z}$ 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 isomorph. 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 $p_n=[p]_n\neq _{nm}$.

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 $p(x,y)\neq (_n,_m)$, which is absurd since $px=_n$ ($p$ is a multiple of $n$) and $py=_m$ ($p$ 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 reccurrence 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 a 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 an 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 suffice 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 an 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*}