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 finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Proof.

Note that as $n$ and $m$ are relatively prime integers, there exits $s, t\in \Z$ such that
\[sn+tm=1. \tag{*}\] Also, note that as the order of the group $G/N$ is $|G/N|=|G:N|=m$, we have
\[g^mN=(gN)^m=N\] for any $g \in G$ by Lagrange’ theorem, and thus
\[g^m\in N \tag{**}.\]

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

Suppose $a\in \{a\in G \mid a^n=e\}$. Then we have $a^n=e$.
It follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=(a^t)^m\in N
\end{align*}
by (**).
This proves that $\{a\in G \mid a^n=e\} \subset N$.

On the other hand, if $a\in N$, then we have $a^n=e$ as $n$ is the order of the group $N$.
Hence $N\subset \{a\in G \mid a^n=e\}$.

Putting together these inclusions yields that $N=\{a\in G \mid a^n=e\}$ as required.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Let $b^m \in \{b^m \mid b\in G\}$. Then by (**), we know that $b^m\in N$.
Thus, we have $\{b^m \mid b\in G\}\subset N$.

On the other hand, let $a\in N$. Then we have $a^n=e$ as $n=|N|$.
Hence it follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=b^m,
\end{align*}
where we put $b:=a^t$.
This implies that $a\in \{b^m \mid b\in G\}$, and hence we have $N \subset \{b^m \mid b\in G\}$.

So we see that $N=\{b^m \mid b\in G\}$ by these two inclusions.

The post Normal Subgroup Whose Order is Relatively Prime to Its Index 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 $G$ be a finite group and let $A, B$ be subsets of $G$ satisfying
\[|A|+|B| > |G|.\] Here $|X|$ denotes the cardinality (the number of elements) of the set $X$.
Then prove that $G=AB$, where
\[AB=\{ab \mid a\in A, b\in B\}.\]

Proof.

Since $A, B$ are subsets of the group $G$, we have $AB\subset G$.
Thus, it remains to show that $G\subset AB$, that is any element $g\in G$ is of the form $ab$ for some $a\in A$ and $b\in B$.
This is equivalent to finding $a\in A$ and $b\in B$ such that $gb^{-1}=a$.

Consider the subset
\[B^{-1}:=\{b^{-1} \mid b \in B\}.\] Since taking the inverse gives the bijective map $B \to B^{-1}$, $b \mapsto b^{-1}$, we have $|B|=|B^{-1}|$.

Also consider the subset
\[gB^{-1}=\{gb^{-1} \mid b\in B\}.\] Note that multiplying by $g$ and by its inverse $g^{-1}$ give the bijective maps
\[B^{-1} \to gB^{-1}, b^{-1} \mapsto gb^{-1} \text{ and } gB^{-1} \to B^{-1}, gb^{-1} \mapsto b^{-1}.\] Hence we have
\[ |B|=|B^{-1}|=|gB^{-1}|.\]

Since $A$ and $gB^{-1}$ are both subsets in $G$ and we have by assumption that
\[|A|+|gB^{-1}|=|A|+|B| > |G|,\] the intersection $A\cap gB^{-1}$ cannot be empty.

Therefore, there exists $a \in A\cap gB^{-1}$, and thus $a\in A$ and $a=gb^{-1}$ for some $b\in B$.
As a result we obtain $g=ab$.
It yields that $G\subset AB$, and we have $G=AB$ as a consequence.

Related Question.

As an application, or use the similar technique, try the following

See the post↴
Each Element in a Finite Field is the Sum of Two Squares
for a proof of this problem.

The post If Two Subsets $A, B$ of a Finite Group $G$ are Large Enough, then $G=AB$ 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.

Problem 448

Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.
The product of $H$ and $N$ is defined to be the subset
\[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\] Prove that the product $H\cdot N$ is a subgroup of $G$.

Definition.

A subgroup $N$ of a group $G$ is called a normal subgroup if for any $g\in G$ and $n\in N$, we have
\[gng^{-1}\in N.\]  

Proof.

We prove that the product $H\cdot N$ is closed under products and inverses.

Let $h_1n_1$ and $h_2n_2$ be elements in $H\cdot N$, where $h_1, h_2\in H$ and $n_1, n_2\in N$.
Let $e$ be the identity element in $G$.
We have
\begin{align*}
(h_1n_1)(h_2n_2)&=h_1en_1h_2n_2\\
&=h_1(h_2h_2^{-1})n_1h_2n_2 && \text{since $h_2h_2^{-1}=e$}\\
&=(h_1h_2)(h_2^{-1}n_1h_2n_2). \tag{*}
\end{align*}
Since $H$ is a subgroup, the element $h_1h_2$ is in $H$.
Also, since $N$ is a normal subgroup, we have $h_2^{-1}n_1h_2$ is in $N$. Hence
\[h_2^{-1}n_1h_2n_2=(h_2^{-1}n_1h_2)n_2\in N.\] It follows from (*) that the product
\[(h_1n_1)(h_2n_2)=(h_1h_2)(h_2^{-1}n_1h_2n_2)\in H\cdot N.\] Therefore, the product $H\cdot N$ is closed under products.

Next, let $hn$ be any element in $H\cdot N$, where $h\in H$ and $n\in N$.
Then we have
\begin{align*}
(hn)^{-1}&=n^{-1}h^{-1}\\
&=en^{-1}h^{-1}\\
&=(h^{-1}h)n^{-1}h^{-1} &&\text{since $h^{-1}h=e$}\\
&=h^{-1}(hn^{-1}h^{-1}).
\end{align*}
Since $N$ is a normal subgroup, we have $hn^{-1}h^{-1}\in N$, and hence
\[(hn)^{-1}=h^{-1}(hn^{-1}h^{-1})\in H\cdot N.\] Thus, the product $H\cdot N$ is closed under inverses.

This completes the proof that the product $H\cdot N$ is a subgroup of $G$.

The post The Product of a Subgroup and a Normal Subgroup is a Subgroup 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.

Let $G$ be a group. Suppose that we have
\[(ab)^3=a^3b^3\] for any elements $a, b$ in $G$. Also suppose that $G$ has no elements of order $3$.

Then prove that $G$ is an abelian group.

Proof.

Let $a, b$ be arbitrary elements of the group $G$. We want to show that $ab=ba$.

By the given relation $(ab)^3=a^3b^3$, we have
\begin{align*}
ababab=a^3b^3.
\end{align*}
Multiplying by $a^{-1}$ on the left and $b^{-1}$ on the right, we obtain
\[baba=a^2b^2,\] or equivalently we have
\[(ba)^2=a^2b^2 \tag{*}\] for any $a, b\in G$.

Now we consider $aba^{-1}b^{-1}$ (such an expression is called the commutator of $a, b$).
We have
\begin{align*}
(aba^{-1}b^{-1})^2&=(a^{-1}b^{-1})^2(ab)^2 && \text{by (*)}\\
&=b^{-2}a^{-2}b^2a^2 && \text{by (*)}\\
&=b^{-2}(ba^{-1})^2a^2 && \text{by (*)}\\
&=b^{-2}ba^{-1}ba^{-1}a^2\\
&=b^{-1}a^{-1}ba.
\end{align*}
Hence we have obtained
\[(aba^{-1}b^{-1})^2=b^{-1}a^{-1}ba \tag{**}\] for any $a, b\in G$.

Taking the square of (**), we obtain
\begin{align*}
(aba^{-1}b^{-1})^4&=(b^{-1}a^{-1}ba)^2\\
&=aba^{-1}b^{-1}. && \text{by (**)}
\end{align*}
It follows that we have
\[(aba^{-1}b^{-1})^3=e,\] where $e$ is the identity element of $G$.

Since the group $G$ does not have an element of order $3$, this yields that
\[aba^{-1}b^{-1}=e.\] (Otherwise, the order of the element $aba^{-1}b^{-1}$ would be $3$.)

This is equivalent to
\[ab=ba.\] Thus, we have obtained $ab=ba$ for any elements $a, b$ in $G$.
Therefore, the group $G$ is abelian.

Related Question.

I came up with this problem when I solved the previous problem:

(For a proof of this problem, see the post “Prove a group is abelian if $(ab)^2=a^2b^2$“.)

I wondered what happens if I change $2$ to $3$, and that’s how I made this problem.

The post Prove a Group is Abelian if $(ab)^3=a^3b^3$ and No Elements of Order $ first appeared on Problems in Mathematics.