In this post, we give selected algebra questions with answers. The exercises focus on group theory and arithmetic. In fact, these types of algebra exercises are classic and students need to learn them.

## A selection of algebra questions with answers

In this section, we shall give various algebra questions with answers.

### An algebraic equation in a quotient set

Let $p$ be an odd prime number and let $(a,b,c)\in\mathbb{Z}^3$ such that $a\notin p\mathbb{Z}$. This means that the class of $a$ denoted by $\overline{a}$ satisfies $\overline{a}\neq \overline{0}$.

- Show that the equation $\overline{a}x^2+\overline{b}x+\overline{c}=\overline{0}$ has solutions in $\mathbb{Z}/p\mathbb{Z}$ if and only if $\Delta=\overline{b^2-4ac}$ is a square in $\mathbb{Z}/p\mathbb{Z}$. In fact, let $x\in \mathbb{Z}/p\mathbb{Z}$. As $\mathbb{Z}/p\mathbb{Z}$ is a field and, $\overline{a}\neq \overline{0}$, and $\overline{2}\neq \overline{0},$ because $p$ is prime, we deduce that $\overline{2}$ and $\overline{a}$ are invertible. We can then write \begin{align*} \overline{a}x^2+\overline{b}x+\overline{c}&= \overline{a}\left( x^2+ \overline{2} (\overline{2}^{-1}\overline{a}^{-1}\overline{b})x\right)+\overline{c}. \end{align*} We select \begin{align*} \overline{d}:=\overline{2}^{-1}\overline{a}^{-1}\overline{b}. \end{align*} Then \begin{align*} \overline{a}x^2+\overline{b}x+\overline{c}&= \overline{a}\left( x^2+ \overline{2} \overline{d} x\right)+\overline{c}\cr &= \overline{a}\left( x+ \overline{d}\right)^2+ \overline{c}- \overline{a}\times \overline{d}^2. \end{align*} Then $x$ is a solution of $\overline{a}x^2+\overline{b}x+\overline{c}$ if and only if \begin{align*} \overline{a}\left( x+ \overline{d}\right)^2&=\overline{a} \overline{d}^2-\overline{c}\cr &= \overline{4a}^{-1}\times \overline{b}-\overline{c}\cr &= \overline{4a}^{-1},\Delta. \end{align*} We conclude that \begin{align*} \Delta= \left(\overline{2a}(x+\overline{d})\right)^2. \end{align*} Finally, the equation has solutions if and only if $\Delta$ is a square in $\mathbb{Z}/p\mathbb{Z}$. Remark that The fact that Whether $\delta$ is a square in $\mathbb{Z}/p\mathbb{Z}$ or not does not depend on the delta sign of $\Delta$. For example, $-1\equiv 2^2\,[5]$ is a square in $\mathbb{Z}/5\mathbb{Z},$ but $3$ does not.
- How many numbers of the solution we have? Determine their expressions in terms of $\Delta$. In fact, From the previous question, solutions exists if $\Delta=\delta^2$ with $\delta\in \mathbb{Z}/p\mathbb{Z}$. We discuss two cases. First, if $\delta=\overline{0},$ in this case we have \begin{align*} \left(\overline{2a}(x+\overline{d})\right)^2=0. \end{align*} So that \begin{align*} x=-\overline{d}=- \overline{2a}^{-1}\times \overline{b}. \end{align*} Second, if $\delta\neq \overline{0},$ it follows that $\overline{2a}(x+\overline{d}=\pm \delta$ and hence \begin{align*} x= \overline{2a}^{-1}(\overline{b}\pm \delta). \end{align*}
- Solve in $\mathbb{Z}/7\mathbb{Z}$ the following equations \begin{align*} x^2+\overline{5}x+\overline{1}=\overline{0},\quad x^2+\overline{2}x+\overline{4}=\overline{0}. \end{align*} In fact, Let consider the equation $x^2+\overline{5}x+\overline{1}=\overline{0}$. In this case we have $\delta=\overline{25}-\overline{4}=\overline{3\times 7}=\overline{0}$ in $\mathbb{Z}/7\mathbb{Z}$. According to the question 2, the unique solution is $x=-\overline{2}\times \overline{5}$. But if we write $5=7-2,$ then $\overline{5}=-\overline{2}$ in $\mathbb{Z}/7\mathbb{Z}$. Thus the solution is $x=-\overline{2}^{-1}\times (-\overline{2})=\overline{1}$. For the second equation $x^2+\overline{2}x+\overline{4}=\overline{0},$ we have $\Delta=\overline{4}-\overline{16}=-\overline{12}=\overline{2}$ in $\mathbb{Z}/7\mathbb{Z}$. Observe that $\overline{9}=\overline{2}$ in $\mathbb{Z}/7\mathbb{Z}$. Then $\Delta=\overline{3}^2$. Thus from question 2, the solutions are $x=\overline{2}^{-1}\times (-\overline{2}\pm \overline{3})$, this means that $x=\overline{2}^{-1}$ or $x=\overline{2}^{-1} \times (-\overline{5})$. As $\overline{2}^{-1}=\overline{4}$ and $-\overline{5}=\overline{2}$ in $\mathbb{Z}/7\mathbb{Z}$, we obtain \begin{align*} x=\overline{4}\quad\text{and}\quad x=\overline{1}. \end{align*}

### Selected algebra exercises on finite group

Let $G$ be a finite abelian group.

- Let $x$ and $y$ be elements of $G$ of orders $p$ and $q$, respectively. Show that if ${\rm gcd}(p,q)=1$ then the order of the element $xy$ is $pq$. In fact, As $G$ is abelian, we have \begin{align*} (xy)^{pq}=x^{pq} y^{pq}= (x^p)^q(y^q)^p=e. \end{align*} Now if we denote by $d$ the order of $xy,$ then $d|pq$. Let us now prove that $pq|d$. In fact, we have $(xy)^d=e$. Using the fact that $G$ is abelian, we obtain $(xy)^{dq}=e$ and then $x^{dq}y^{dq}=e$. As $y^{dq}=e$, then $x^{dq}=e$. As $p$ is the order of $x,$ $p|pq$. But $gcd(p,q)=1$, so that $p|d$. Using similar argent and the fact that $(xy)^{dp}=e$, we obtain $q|d$. This implies that $pq|d$ because $gcd(p,q)=1$. Hence $d=pq$.
- Deduce that there exists $x\in G$ with an order equal to the lowest common multiple, “lcm” of elements of $G$. In fact, let us prove that for any $i=1,\cdots,s$, there exists $x_i$ element of $G$ of order $p_{i}^{\alpha_i}$. In fact, let $i\in {1,\cdots,s}$ and denote by $A$ the set of all orders of elements of $G$. As $m$ is the lowest common multiple of elements of $G$, we have $\alpha_i=nu_{p_i}(m)=\max_{d\in A}nu_{p_i}(d),$ here $\nu_{p_i}$ is the $p_i$-adic valuation. Hence $y\in G$ of order $d$ satisfies $nu_{p_i}(d)=\alpha_i$. So $p_{i}^{\alpha_i}|d$ and there exists $\ell\in\mathbb{N}$ such that $d=p_{i}^{\alpha_i}\ell$. This implies that the element $x_i=y^{\ell}$ is of order $p_{i}^{\alpha_i}$. We now select \begin{align*} x=x_1\cdots x_s. \end{align*} By using a recurrence argument, one can see that $x$ is of order $m$.

The subject is beyond the scope of the first course in groups.