Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.

(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?

(b) Let $ax+b+I$ be a nonzero element of the field $\F_3[x]/(x^2+1)$, where $a, b \in \F_3$. Find the inverse of $ax+b+I$.

(c) Recall that the multiplicative group of nonzero elements of a field is a cyclic group.

Confirm that the element $x$ is not a generator of $E^{\times}$, where $E=\F_3[x]/(x^2+1)$ but $x+1$ is a generator.

Proof.

(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field

Let $f(x)=x^2+1$. We claim that the polynomial $f(x)$ is irreducible over $\F_3$.
To see this, note that $f(x)$ is a quadratic polynomial.
So $f(x)$ is irreducible over $\F_3$ if it does not have a root in $\F_3$.
We have
\begin{align*}
f(0)=1, \quad f(1)=2, \quad f(2)=2^2+1=2 \text{ in } \F_3.
\end{align*}
Hence $f(x)$ does not have a root in $\F_3$ and it is irreducible over $\F_3$.

It follows that the quotient $\F_3[x]/(x^2+1)$ is a field.
Since $x^2+1$ is quadratic, the extension degree of $\F_3[x]/(x^2+1)$ over $\F_3$ is $2$.
Hence the number of elements in the field is $3^2=9$.

(b) Find the inverse of $ax+b+I$

Let $ax+b$ be a representative of a nonzero element of the field $\F_3[x]/(x^2+1)$.
Let $cx+d$ be its inverse. Then we have
\begin{align*}
1&=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd\\
&=(ad+bc)x+bd-ac
\end{align*}
since $x^2=-1$ in $\F_3[x]/(x^2+1)$.

Hence we obtain two equations
\begin{align*}
ad+bc=0 \text{ and } bd-ac=1.
\end{align*}

Since $ax+b$ is a nonzero element, at least one of $a, b$ is not zero.
If $a\neq 0$, then the first equation gives
\[d=-\frac{bc}{a}. \tag{*}\] Substituting this to the second equation, we obtain
\begin{align*}
\left(\, \frac{-b^2-a^2}{a} \,\right)c=1.
\end{align*}
Observe that $a^2+b^2$ is not zero in $\F_3$.
(Since $a \neq 0$, we have $a^2=1$. Also $b^2=0, 1$.)
Hence we have
\begin{align*}
c=-\frac{a}{a^2+b^2}.
\end{align*}

It follows from (*) that
\[d=\frac{b}{a^2+b^2}\]

Thus, if $a \neq 0$, then the inverse element is
\[(ax+b)^{-1}=\frac{1}{a^2+b^2}(-ax+b). \tag{**}\]

If $a=0$, then $b\neq 0$ and it is clear that the inverse element of $ax+b=b$ is $1/b$.
Note that the formula (**) is still true in this case.

In summary, we have

for any nonzero element $ax+b$ in the field $\F_3[x]/(x^2+1)$.

(c) $x$ is not a generator but $x+1$ is a generator

Note that the order of $E^{\times}$ is $8$ since $E$ is a finite field of order $9$ by part (a).
We compute the powers of $x$ and obtain
\begin{align*}
x, \quad x^2=-1, \quad x^3=-x, \quad x^4=-x^2=1.
\end{align*}
Thus, the order of the element $x$ is $4$, hence $x$ is not a generator of the cyclic group $E^{\times}$.

Next, let us check that $x+1$ is a generator.
We compute the powers of $x+1$ as follows.
\begin{align*}
&x+1, \quad (x+1)^2=x^2+2x+1=2x, \\
&(x+1)^3=2x(x+1)=2x^2+2x=2x-2=2x+1\\
&(x+1)^4=(2x+1)(x+1)=2x^2+3x+1=2.
\end{align*}

Observe that at this post the order of $x+1$ must be larger than $4$.
Since the order of $E^{\times}$ is $8$, the order of $x+1$ must be $8$ by Lagrange’s theorem.

Just for a reference we give the complete list of powers of $x+1$.
\[\begin{array}{ |c|c|}
\hline
n & (x+1)^n \\
\hline
1 & x+1 \\
2 & 2x \\
3 & 2x+1 \\
4 & 2 \\
5 & 2x+2\\
6 & x\\
7 &x+2\\
8 & 1\\
\hline
\end{array}\] Click here if solved 59

The post Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements first appeared on Problems in Mathematics.

Let $G, G’$ be groups. Let $\phi:G\to G’$ be a group homomorphism.
Then prove that for any element $g\in G$, we have
\[\phi(g^{-1})=\phi(g)^{-1}.\]

Definition (Group homomorphism).

A map $\phi:G\to G’$ is called a group homomorphism if
\[\phi(ab)=\phi(a)\phi(b)\] for any elements $a, b\in G$.

Proof.

Let $e, e’$ be the identity elements of $G, G’$, respectively.
First we claim that
\[\phi(e)=e’.\] In fact, we have
\begin{align*}
\phi(e)&=\phi(ee)=\phi(e)\phi(e) \tag{*}
\end{align*}
since $\phi$ is a group homomorphism.
Thus, multiplying by $\phi(e)^{-1}$ on the left, we obtain
\begin{align*}
&e’=\phi(e)^{-1}\phi(e)\\
&=\phi(e)^{-1}\phi(e)\phi(e) && \text{by (*)}\\
&=e’\phi(e)=\phi(e).
\end{align*}
Hence the claim is proved.

Then we have
\begin{align*}
&e’=\phi(e) && \text{by claim}\\
&=\phi(gg^{-1})\\
&=\phi(g)\phi(g^{-1}) && \text{since $\phi$ is a group homomorphism}.
\end{align*}

It follows that we have
\begin{align*}
\phi(g)^{-1}&=\phi(g)^{-1}e’\\
&=\phi(g)^{-1}\phi(g)\phi(g^{-1})\\
&=e’\phi(g^{-1})\\
&=\phi(g^{-1}).
\end{align*}
This completes the proof.

The post Group Homomorphism Sends the Inverse Element to the Inverse Element first appeared on Problems in Mathematics.

Prove that if $G$ is a finite group of even order, then the number of elements of $G$ of order $2$ is odd.

Proof.

First observe that for $g\in G$,
\[g^2=e \iff g=g^{-1},\] where $e$ is the identity element of $G$.
Thus, the identity element $e$ and the elements of order $2$ are the only elements of $G$ that are equal to their own inverse elements.

Hence, each element $x$ of order greater than $2$ comes in pairs $\{x, x^{-1}\}$.
So we have
\begin{align*}
&G=\\
&\{e\}\cup \{\text{ elements of order $2$ } \} \cup \{x_1, x_1^{-1}, x_2, x_2^{-1}, \dots, x_k, x_k^{-1}\},
\end{align*}
where $x_i$ are elements of order greater than $2$ for $i=1,2, \dots, k$.

As we noted above, the elements $x_i, x_i^{-1}$ are distinct.
Thus the third set contains an even number of elements.

Therefore we have
\begin{align*}
&\underbrace{G}_{\text{even}}=\\
&\underbrace{\{e\}}_{\text{odd}}\cup \{\text{ elements of order $2$ } \}\cup \underbrace{\{x_1, x_1^{-1}, x_2, x_2^{-1}, \dots, x_k, x_k^{-1}\}}_\text{even}
\end{align*}
It follows that the number of elements of $G$ of order $2$ must be odd.

If the Order of a Group is Even, then it has a Non-Identity Element of Order 2

The consequence of the problem yields that the number of elements of order $2$ is odd, in particular, it is not zero.

Hence we obtain:

The post If the Order of a Group is Even, then the Number of Elements of Order 2 is Odd first appeared on Problems in Mathematics.

Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.

(a) Prove that $T(A)$ is a subgroup of $A$.

(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion elements.)

(b) Prove that the quotient group $G=A/T(A)$ is a torsion-free abelian group. That is, the only element of $G$ that has finite order is the identity element.

Proof.

(a) $T(A)$ is a subgroup of $A$

We write the group operation multiplicatively.
Let $x, y\in T(A)$. Then $x, y$ have finite order, hence there exists positive integers $m, n$ such that $x^m=e, y^n=e$, where $e$ is the identity element of $A$. Then we have
\begin{align*}
(xy)^{mn}&=x^{mn}y^{mn} \qquad \text{ (since $A$ is abelian)}\\
&=(x^m)^n(y^m)^n=e^me^n=e.
\end{align*}
Therefore the element $xy$ has also finite order, hence $xy \in T(A)$.

Also, we have
\begin{align*}
(x^{-1})^m=(x^m)^{-1}=e^{-1}=e.
\end{align*}
Hence the inverse $x^{-1}$ of $x$ has finite order, hence $x^{-1}\in T(A)$.

Therefore, the subset $T(A)$ is closed under group operation and inverse, hence $T(A)$ is a subgroup of $A$.

(b) $A/T(A)$ is a torsion-free abelian group

Since $A$ is an abelian group, the quotient $G=A/T(A)$ is also an abelian group.
For $a\in A$, let $\bar{a}=aT(A)$ be an element of $G=A/T(A)$. Suppose that $\bar{a}$ has finite order in $G$. We want to prove that $\bar{a}=\bar{e}$ the identity element of $G$.

Since $\bar{a}$ has finite order, there exists a positive integer $n$ such that
\[\bar{a}^n=\bar{e}.\] This implies that
\[a^nT(A)=T(A)\] and thus $a^n\in T(A)$.

Since each element of $T(A)$ has finite order by definition, there exists a positive integer $m$ such that $(a^n)^m=e$.
It follows from $a^{nm}=e$ that $a$ has finite order, and thus $a\in T(A)$.
Therefore we have
\[\bar{a}=aT(A)=T(A)=\bar{e}.\]

We have proved that any element of $G=A/T(A)$ that has finite order is the identity, hence $G$ is the torsion-free abelian subgroup of $G$.

The post Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group first appeared on Problems in Mathematics.

Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.
Define the subset $M$ of $G_1 \times G_2$ to be
\[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\]

Prove that $M$ is a subgroup of $G_1 \times G_2$.

Proof.

$M$ is closed under the group operation

Suppose that $(a_1, a_2), (b_1, b_2) \in M$.
By definition, we have
\[f_1(a_1)=f_2(a_2) \text{ and } f_1(b_1)=f_2(b_2). \tag{*}\] The product of $(a_1, a_2)$ and $(b_1, b_2)$ is
\[(a_1, a_2)\cdot (b_1, b_2)=(a_1b_1, a_2b_2).\]

We want to prove that this is in $M$. To see this, note that we have
\begin{align*}
f_1(a_1b_1)&=f_1(a_1)f_1(b_1) && \text{($f_1$ is a homomorphism)}\\
&=f_2(a_2)f_2(b_2) && \text{(by (*))}\\
&=f_2(a_2b_2) && \text{($f_2$ is a homomorphism)}.
\end{align*}

Therefore we have obtained
\[f_1(a_1b_1)=f_2(a_2b_2),\] and the product $(a_1b_1, a_2b_2)$ is in $M$ by definition.

$M$ is closed under inverses

We next prove that if $(a_1, a_2)\in M$ then the inverse
\[(a_1, a_2)^{-1}=(a_1^{-1}, a_2^{-1})\] is also in $M$.

Since $(a_1, a_2)\in M$, we have
\[f_1(a_1)=f_2(a_2). \tag{**}\] Then we have
\begin{align*}
f_1(a_1^{-1})&=(f_1(a_1))^{-1} && \text{($f_1$ is a homomorphism)}\\
&=(f_2(a_2))^{-1} && \text{ (by (**))}\\
&=f_2(a_2^{-1}) && \text{($f_2$ is a homomorphism)}.
\end{align*}
Thus by definition the inverse $(a_1^{-1}, a_2^{-1})$ is in $M$.

Since $M$ is closed under the group operation and inverses, it is a subgroup of $G_1 \times G_2$.

Comment.

In category theory, the subgroup $M$ in the problem is called a pullback of homomorphisms $f_1: G_1 \to H$ and $f_2: G_2 \to H$.

The post Pullback Group of Two Group Homomorphisms into a Group first appeared on Problems in Mathematics.

Let $G$ be a group. Define a map $f:G \to G$ by sending each element $g \in G$ to its inverse $g^{-1} \in G$.
Show that $G$ is an abelian group if and only if the map $f: G\to G$ is a group homomorphism.

Proof.

$(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism.

Suppose that $G$ is an abelian group. Then we have for any $g, h \in G$
\begin{align*}
f(gh)&=(gh)^{-1}=h^{-1}g^{-1}\\
&=g^{-1}h^{-1} \text{ since } G \text{ is abelian}\\
&=f(g)f(h).
\end{align*}
This implies that the map $f$ is a group homomorphism.

$(\impliedby)$ If $f$ is a homomorphism, then $G$ is an abelian group.

Now we suppose that the map $f: G \to G$ is a group homomorphism.
Then for any $g, h \in G$, we have
\[f(gh)=f(g)f(h) \tag{*}\] since $f$ is a group homomorphism.
The left hand side of (*) is
\[f(gh)=(gh)^{-1}=h^{-1}g^{-1}.\] Thus we obtain from (*) that
\[h^{-1}g^{-1}=g^{-1}h^{-1}.\] Taking the inverse of both sides, we have
\[gh=hg\] for any $g, h \in G$.
It follows that $G$ is an abelian group.

Related Question.

Another problem about the relation between an abelian group and a group homomorphism is:
A group is abelian if and only if squaring is a group homomorphism

The post A Group Homomorphism and an Abelian Group first appeared on Problems in Mathematics.