Home Post Selected algebra exercises for college students

# Selected algebra exercises for college students

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

### An algebraic equation in a quotient set

Let $p$ be an odd prime number and let $(a,b,c)inmathbb{Z}^3$ such that $anotin pmathbb{Z}$ (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 $mathbb{Z}/pmathbb{Z}$ if and only if $Delta=overline{b^2-4ac}$ is a square in $mathbb{Z}/pmathbb{Z}$. In fact, let $xin mathbb{Z}/pmathbb{Z}$. As $mathbb{Z}/pmathbb{Z}$ 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})xright)+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} xright)+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 have solutions if and only if $Delta$ is a square in $mathbb{Z}/pmathbb{Z}$. Remark that The fact that Whether $delta$ is a square in $mathbb{Z}/pmathbb{Z}$ or not does not depend on the delta sign of $Delta$. For example $-1equiv 2^2,[5]$ is a square in $mathbb{Z}/5mathbb{Z},$ 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 $deltain mathbb{Z}/pmathbb{Z}$. 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 $deltaneq overline{0},$ 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 $mathbb{Z}/7mathbb{Z}$ 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 $delta=overline{25}-overline{4}=overline{3times 7}=overline{0}$ in $mathbb{Z}/7mathbb{Z}$. 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 $mathbb{Z}/7mathbb{Z}$. 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 $mathbb{Z}/7mathbb{Z}$. Observe that $overline{9}=overline{2}$ in $mathbb{Z}/7mathbb{Z}$. 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 $mathbb{Z}/7mathbb{Z}$, we obtain begin{align*} x=overline{4}quadtext{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 $xin G$ with 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 $iin {1,cdots,s}$ 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 $alpha_i=nu_{p_i}(m)=max_{din A}nu_{p_i}(d)$ (here $nu_{p_i}$ is the $p_i$-adic valuation). Hence $yin G$ of order $d$ satisfies $nu_{p_i}(d)=alpha_i$. So $p_{i}^{alpha_i}|d$ and there exists $ellinmathbb{N}$ 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_1cdots x_s. end{align*} By using a recurrence argument, one can see that $x$ is of order $m$.
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