Problem 628

Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.

Then determine the number of elements in $G$ of order $3$.
 

Proof.

Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.

By Sylow’s theorem, we know that
\[n_{19} \equiv 1 \pmod{19} \text{ and } n_{19} \mid 3.\] It follows that $n_{19}=1$.

Now, observe that if $g\in G$, then the order of $g$ is $1$, $3$, or $19$. Note that since $G$ is not a cyclic group, the order of $g$ cannot be $57$.
As there is exactly one Sylow $19$-subgroup $P$, any element that is not in $P$ must have order $3$.

Therefore, the number of elements of order $3$ is $57-19=38$.

Remark.

Note that there are $18$ elements of order $19$ and the identity element is the only element of order $1$.

The post Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57 first appeared on Problems in Mathematics.

Let $G$ be a group. Suppose that the number of elements in $G$ of order $5$ is $28$.

Determine the number of distinct subgroups of $G$ of order $5$.

Solution.

Let $g$ be an element in $G$ of order $5$.
Then the subgroup $\langle g \rangle$ generated by $g$ is a cyclic group of order $5$.
That is, $\langle g \rangle=\{e, g, g^2, g^3, g^4\}$, where $e$ is the identity element in $G$.

Note that the order of each non-identity element in $\langle g \rangle$ is $5$.

Also, if $h$ is another element in $G$ of order $5$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle = \{e\}$.
This follows from the fact that the intersection $\langle g \rangle \cap \langle h \rangle$ is a subgroup of the order $5$ group $\langle g \rangle$, and thus the order of $\langle g \rangle \cap \langle h \rangle$ is either $5$ or $1$.

On the other hand, if $H$ is a subgroup of $G$ of order $5$, then every non-identity element in $H$ has order $5$.

These observations imply that each subgroup of order $5$ contains exactly $4$ elements of order $5$ and each element of order $5$ appears in exactly one of such subgroups.

As there are $28$ elements of order $5$, there are $28/4=7$ subgroups of order $5$.

The post If There are 28 Elements of Order 5, How Many Subgroups of Order 5? first appeared on Problems in Mathematics.

Prove that every cyclic group is abelian.

Proof.

Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.)

Let $a$ and $b$ be arbitrary elements in $G$.
Then there exists $n, m\in \Z$ such that $a=g^n$ and $b=g^m$.

It follows that
\begin{align*}
ab&=g^ng^m=g^{n+m}=g^mg^n=ba.
\end{align*}

Hence we obtain $ab=ba$ for arbitrary $a, b\in G$.
Thus $G$ is an abelian group.

The post Every Cyclic Group is Abelian first appeared on Problems in Mathematics.

Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is the identity element in the group $G$.

Prove that the number of elements in $S$ is odd.

Proof.

Let $g\neq e$ be an element in the group $G$ such that $g^5=e$.
As $5$ is a prime number, this yields that the order of $g$ is $5$.

Consider the subgroup $\langle g \rangle$ generated by $g$.
As the order of $g$ is $5$, the order of the subgroup $\langle g \rangle$ is $5$.

If $h\neq e$ is another element in $G$ such that $h^5=e$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle=\{e\}$ as the intersection of these two subgroups is a subgroup of $\langle g \rangle$.

It follows that $S$ is the union of subgroups of order $5$ that intersect only at the identity element $e$.
Thus the number of elements in $S$ are $4n+1$ for some nonnegative integer $n$.

The post The Number of Elements Satisfying $g^5=e$ in a Finite Group is Odd first appeared on Problems in Mathematics.

Let $m$ and $n$ be positive integers such that $m \mid n$.

(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

Proof.

(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

To show that $\phi$ is well-defined, we need show that the value of $\phi$ does not depends on the choice of representative $a$.
So suppose that $a+n\Z=a’+n\Z$ so that $a$ and $a’$ are two representatives for the same element.
This yields that $a-a’$ is divisible by $n$.

Now, $a+n\Z$ is mapped to $a+m\Z$ by $\phi$. On the other hand, $a’+n\Z$ is mapped to $a+m\Z$ by $\phi$.
Since $a-a’$ is divisible by $n$ and $m \mid n$, it follows that $a-a’$ is divisible by $m$.
This implies that $a+m\Z=a’+m\Z$.
This prove that $\phi$ does not depend on the choice of the representative, and hence $\phi$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

Let $a+n\Z$, $b+n\Z$ be two elements in $\Zmod{n}$. Then we have
\begin{align*}
&\phi\left(\, (a+n\Z)+(b+n\Z) \,\right)\\
&=\phi\left(\, (a+b)+n\Z) \,\right) &&\text{by addition in $\Zmod{n}$}\\
&=(a+b)+m\Z &&\text{by definition of $\phi$}\\
&=(a+m\Z)+(b+m\Z)&&\text{by addition in $\Zmod{m}$}\\
&=\phi(a+n\Z)+\phi(b+n\Z) &&\text{by definition of $\phi$}.
\end{align*}

Hence $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

For any $c+m\Z \in \Zmod{m}$, we pick $c+n\Z\in \Zmod{n}$.
Then as $\phi(c+n\Z)=c+m\Z$, we see that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

If $a+n\Z\in \ker(\phi)$, then we have $0+m\Z=\phi(a+n\Z)=a+m\Z$.
This implies that $m\mid a$.
On the other hand, if $m\mid a$, then $\phi(a+n\Z)=a+m\Z=0+m\Z$ and $a+n\Z\in \ker(\phi)$.

It follows that
\[\ker(\phi)=\{mk+n\Z \mid k=0, 1, \dots, l-1\},\] where $l$ is an integer such that $n=ml$.

Thus, $\ker(\phi)$ is a group of order $l$.
Since $\ker(\phi)$ is a subgroup of the cyclic group $\Zmod{n}$, we know that $\ker(\phi)$ is also cyclic.
Thus
\[\ker(\phi)\cong \Zmod{l}.\]

Another approach

Here is a more direct proof of this result.
Define a map $\psi:\Z\to \ker(\phi)$ by sending $k\in \Z$ to $mk+n\Z$.
It is straightforward to verify that $\psi$ is a surjective group homomorphism and the kernel of $\psi$ is $\ker(\psi)=l\Z$.
It follows from the first isomorphism theorem that
\[\Zmod{l}= \Z/\ker(\psi) \cong \im(\psi)=\ker(\phi). \] Click here if solved 125

The post Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$ first appeared on Problems in Mathematics.

Let $N$ be a normal subgroup of a group $G$.
Suppose that $G/N$ is an infinite cyclic group.

Then prove that for each positive integer $n$, there exists a normal subgroup $H$ of $G$ of index $n$.

Hint.

Use the fourth (or Lattice) isomorphism theorem.

Proof.

Let $n$ be a positive integer.
Since $G/N$ is a cyclic group, let $g$ be a generator of $G/N$.
So we have $G/N=\langle g\rangle$.
Then $\langle g^n \rangle$ is a subgroup of $G/N$ of index $n$.

By the fourth isomorphism theorem, every subgroup of $G/N$ is of the form $H/N$ for some subgroup $H$ of $G$ containing $N$.
Thus we have $\langle g^n \rangle=H/N$ for some subgroup $H$ in $G$ containing $N$.

Since $G/N$ is cyclic, it is in particular abelian.
Thus $H/N$ is a normal subgroup of $G/N$.

The fourth isomorphism theorem also implies that $H$ is a normal subgroup of $G$, and we have
\begin{align*}
[G:H]=[G/N : H/N]=n.
\end{align*}
Hence $H$ is a normal subgroup of $G$ of index $n$.

The post If the Quotient is an Infinite Cyclic Group, then Exists a Normal Subgroup of Index $n$ first appeared on Problems in Mathematics.

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 $R$ be a ring with unit $1$. Suppose that the order of $R$ is $|R|=p^2$ for some prime number $p$.
Then prove that $R$ is a commutative ring.

Proof.

Let us consider the subset
\[Z:=\{z\in R \mid zr=rz \text{ for any } r\in R\}.\] (This is called the center of the ring $R$.)

This is a subgroup of the additive group $R$.
In fact, if $z, z’\in Z$, then we have for any $r\in R$,
\begin{align*}
(z-z’)r=zr-z’r=rz-rz’=r(z-z’).
\end{align*}
It follows that $z-z’\in Z$, and thus $Z$ is a subgroup of $R$.

Note that $0, 1 \in Z$, hence $Z$ is not a trivial subgroup.
Thus, we have either $|Z|=p, p^2$ since $R$ is a group of order $p^2$.

If $|Z|=p^2$, then we have $Z=R$.
By definition of $Z$, this implies that $R$ is commutative.

It remains to show that $|Z|\neq p$.
Assume that $|Z|=p$.
Then $R/Z$ is a cyclic group of order $p$.
Let $\alpha$ be a generator of $R/Z$.

Since $Z\neq R$, there exist $r, s\in R$ such that $rs\neq sr$.
Write
\[r=m\alpha+z \text{ and } s=n\alpha+z’\] for some $m, n\in \Z$, $z, z’\in Z$.

Then we have
\begin{align*}
rs&=(m\alpha+z)(n\alpha+z’)\\
&=(m\alpha)(n\alpha)+m\alpha z’ + n z\alpha +z z’\\
&=(n\alpha)(m\alpha)+m z’ \alpha +n \alpha z +z’ z\\
&=(n\alpha+z’)(m\alpha+z)\\
&=sr.
\end{align*}

This contradicts $rs\neq sr$, and we conclude that $|Z|\neq p$.

The post Every Ring of Order $p^2$ is Commutative first appeared on Problems in Mathematics.

Let $\Q=(\Q, +)$ be the additive group of rational numbers.

(a) Prove that every finitely generated subgroup of $(\Q, +)$ is cyclic.

(b) Prove that $\Q$ and $\Q \times \Q$ are not isomorphic as groups.

Proof.

(a) Prove that every finitely generated subgroup of $(\Q, +)$ is cyclic.

Let $G$ be a finitely generated subgroup of $(\Q, +)$ and let $r_1, \dots, r_n$ be nonzero generators of $G$.
Let us express
\[r_i=\frac{a_i}{b_i},\] where $a_i, b_i$ are integers.

Let
\[s:=\frac{1}{\prod_{j=1}^n b_j} \in \Q.\] Then we can write each $r_i$ as
\[r_i=\frac{a_i}{b_i}=\left(\, a_i\prod_{\substack{j=1\\j\neq i}}^n b_i \,\right)\cdot \frac{1}{s}.\]

It follows from the last expressions that the elements $r_i$ is contained in the subgroup $\langle s \rangle$ generated by the element $s$.
Hence $G$ is a subgroup of $\langle s \rangle$.
Since every subgroup of a cyclic group is cyclic, we conclude that $G$ is also cyclic.

(b) Prove that $\Q$ and $\Q \times \Q$ are not isomorphic as groups.

Seeking a contradiction, assume that $\Q$ is isomorphic to the direct product $\Q \times \Q$:
\[\Q\cong \Q\times \Q.\]

Then consider the subgroup $\Z\times \Z$ of $\Q\times \Q$.
We claim that the subgroup $\Z\times \Z$ is not cyclic.
If it were cyclic, then there would be a generator $(a,b)\in \Z\times \Z$.

However, for example, the element $(b, -a)$ cannot be expressed as an integer multiple of $(a, b)$.
To see this, suppose that
\[n(a,b)=(b,-a)\] for some integer $n$.

Then we have $na=b$ and $nb=-a$. Substituting the first equality into the second one, we obtain
\[n^2a=-a.\] If $a\neq 0$, then this yields that $n^2=-1$, which is impossible, and hence $a=0$.

Then $na=b$ implies $b=0$ as well.
However, $(a,b)=(0,0)$ is clearly not a generator of $\Z\times \Z$.

Thus we have reached a contradiction and $\Z\times \Z$ is a non-cyclic subgroup of $\Q\times \Q$.
This implies via the isomorphism $\Q\cong \Q \times \Q$ that $\Q$ has a non-cyclic subgroup.
We saw in part (a) that this is impossible.
Therefore, $\Q$ is not isomorphic to $\Q\times \Q$.

The post Every Finitely Generated Subgroup of Additive Group $\Q$ of Rational Numbers is Cyclic first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $217$.

(a) Prove that $G$ is a cyclic group.

(b) Determine the number of generators of the group $G$.

Sylow’s Theorem

We will use Sylow’s theorem to prove part (a).

For a review of Sylow’s theorem, check out the post “Sylow’s Theorem (summary)“.

Proof.

(a) Prove that $G$ is a cyclic group.

Note the prime factorization $217=7\cdot 31$.
We first determine the number $n_p$ of Sylow $p$-group for $p=7, 31$.
Recall from Sylow’s theorem that
\begin{align*}
&n_p \equiv 1 \pmod{p}\\[6pt] &n_p \text{ divides } n/p.
\end{align*}

Thus, $n_7$ could be $1, 8, 15, 22, 29,\dots$ and $n_7$ needs to divide $217/7=31$.
Hence the only possible value for $n_7$ is $n_7=1$.
So there is a unique Sylow $7$-subgroup $P_7$ of $G$.

By Sylow’s theorem, the unique Sylow $7$-subgroup must be a normal subgroup of $G$.

Similarly, $n_{31}=1, 32, \dots$ and $n_{31}$ must divide $217/31=7$, and hence we must have $n_{31}=1$.
Thus $G$ has a unique normal Sylow $31$-subgroup $P_{31}$.

Note that these Sylow subgroup have prime order, and hence they are isomorphic to cyclic groups:
\[P_7\cong \Zmod{7} \text{ and } P_{31}\cong \Zmod{31}.\]

It is also straightforward to see that $P_7 \cap P_{31}=\{e\}$, where $e$ is the identity element in $G$.

In summary, we have

1. $P_7, P_{31}$ are normal subgroups of $G$.
2. $P_7 \cap P_{31}=\{e\}$.
3. $|P_7P_{31}|=|G|$.

These yields that $G$ is a direct product of $P_7$ and $P_{31}$, and we obtain
\[G=P_7\times P_{31}\cong \Zmod{7} \times \Zmod{31}\cong \Zmod{217}.\] Hence $G$ is a cyclic group.

(b) Determine the number of generators of the group $G$.

Recall that the number of generators of a cyclic group of order $n$ is equal to the number of integers between $1$ and $n$ that are relatively prime to $n$.
Namely, the number of generators is equal to $\phi(n)$, where $\phi$ is the Euler totient function.

By part (a), we know that $G$ is a cyclic group of order $217$.
Thus, the number of generators of $G$ is
\begin{align*}
\phi(217)=\phi(7)\phi(31)=6\cdot 30=180,
\end{align*}
where the first equality follows since $\phi$ is multiplicative.

The post Prove that a Group of Order 217 is Cyclic and Find the Number of Generators first appeared on Problems in Mathematics.