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.

Let $G$ a finite group and let $H$ and $K$ be two distinct Sylow $p$-group, where $p$ is a prime number dividing the order $|G|$ of $G$.

Prove that the product $HK$ can never be a subgroup of the group $G$.

Hint.

Use the following fact.

Proof.

Let $p^{\alpha}$ is the highest power of $p$ that divides $|G|$.
That is, we have
\[|G|=p^{\alpha}n,\] where $p$ does not divide the integer $n$.

Then the orders of the Sylow $p$-subgroups $H, K$ are $p^{\alpha}$.

Since the intersection $H\cap K$ is a subgroup of $H$, the order of $H \cap K$ is $p^{\beta}$ for some integer $\beta \leq \alpha$ by Lagrange’s theorem.
As $H$ and $K$ are distinct subgroups, we must have $\beta < \alpha$.

Then the number of elements of the product $HK$ is
\begin{align*}
|HK|&=\frac{|H| |K|}{|H \cap K|}\\[6pt] &=\frac{p^{\alpha} p^{\alpha}}{p^{\beta}}=p^{2\alpha-\beta}.
\end{align*}
Since $\beta < \alpha$, we have $2\alpha-\beta > \alpha$.

It follows that the product $HK$ cannot be a subgroup of $G$ since otherwise the order $|HK|=p^{2\alpha-\beta}$ divides $|G|$ by Lagrange’s theorem but $p^{\alpha}$ is the highest power of $p$ that divides $G$.

The post The Product of Distinct Sylow $p$-Subgroups Can Never be a Subgroup first appeared on Problems in Mathematics.

Let $a, b$ be relatively prime integers and let $p$ be a prime number.
Suppose that we have
\[a^{2^n}+b^{2^n}\equiv 0 \pmod{p}\] for some positive integer $n$.

Then prove that $2^{n+1}$ divides $p-1$.

Proof.

Since $a$ and $b$ are relatively prime, at least one of them is relatively prime to $p$.
Without loss of generality let us assume that $b$ and $p$ are relatively prime.

Then the given equality becomes
\begin{align*}
a^{2^n}\equiv -b^{2^n} \pmod{p} \\
\iff \left( \frac{a}{b}\right)^{2^n} \equiv -1 \pmod{p}.
\end{align*}
Taking square of both sides we obtain
\[\left( \frac{a}{b}\right)^{2^{n+1}} \equiv 1 \pmod{p}.\]

Now, we can think of these congruences as equalities of elements in the multiplicative group $(\Z/p\Z)^{\times}$ of order $p-1$:
\[ \left( \frac{a}{b}\right)^{2^n} = -1 \text{ and } \left( \frac{a}{b}\right)^{2^{n+1}} =1 \text{ in } (\Z/p\Z)^{\times}.\]

Note that the second equality yields that the order of the element $a/b$ divides $2^{n+1}$.
On the other hand, the first equality implies that any smaller power of $2$ is not the order of $a/b$.
Thus, the order of the element $a/b$ is exactly $2^{n+1}$.

In general, the order of each element divides the order of the group.
(This is a consequence of Lagrange’s theorem.)

Since the order of the group $(\Z/p\Z)^{\times}$ is $p-1$, it follows that $2^{n+1}$ divides $p-1$.
This completes the proof.

The post Number Theoretical Problem Proved by Group Theory. $a^{2^n}+b^{2^n}\equiv 0 \pmod{p}$ Implies ^{n+1}|p-1$. first appeared on Problems in Mathematics.

A nontrivial abelian group $A$ is called divisible if for each element $a\in A$ and each nonzero integer $k$, there is an element $x \in A$ such that $x^k=a$.
(Here the group operation of $A$ is written multiplicatively. In additive notation, the equation is written as $kx=a$.) That is, $A$ is divisible if each element has a $k$-th root in $A$.

(a) Prove that the additive group of rational numbers $\Q$ is divisible.

(b) Prove that no finite abelian group is divisible.

Proof.

(a) The additive group of rational numbers $\Q$ is divisible.

We know that $\Q$ is a nontrivial abelian group.
Let $a\in Q$ and $k$ be a nonzero integer.
Since $\Q$ is an additive group, we are seeking $x\in \Q$ such that
\[kx=a.\]

Since $k$ is a nonzero integer, we have a solution $x=a/k\in \Q$.
Thus $\Q$ is divisible.

(b) No finite abelian group id divisible.

Let $G$ be a finite abelian group of order $|G|=n$.
If $G$ is trivial, that is, $n=1$, then by definition, $G$ is not divisible.
So let us assume that $G$ is nontrivial.

We claim that $G$ is not divisible since there is no $n$-th root of a nonidentity element of $G$.
Let $a\in G$ be a nonidentity element of $G$.
(Such an element exists because $G$ is nontrivial.)

If there is $x\in G$ such that
\[x^n=a,\] then by Lagrange’s theorem we have $x^n=e$, the identity element of $G$.
This implies that $a=e$, and this contradicts our choice of $a$.
Thus $G$ is not divisible.

The post No Finite Abelian Group is Divisible first appeared on Problems in Mathematics.

Use Lagrange’s Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat’s Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.

 

Before the proof, let us recall Lagrange’s Theorem.

Lagrange’s Theorem

If $G$ is a finite group and $H$ is a subgroup of $G$, then the order $|H|$ of $H$ divides the order $|G|$ of $G$.

Proof.

If $a=0$, then we clearly have $a^p \equiv a \pmod p$.
So we assume that $a\neq 0$.
Then $\bar{a}=a+p\Z \in (\Zmod{p})^{\times}$.

Let $H$ be a subgroup of $(\Zmod{p})^{\times}$ generated by $\bar{a}$.
Then the order of the subgroup $H$ is the order of the element $\bar{a}$.

By Lagrange’s Theorem, the order $|H|$ divides the order of the group $(\Zmod{p})^{\times}$, which is $p-1$.
So we write $p-1=|H|m$ for some $m \in \Z$.

Therefore, we have
\begin{align*}
\bar{a}^{p-1}=\bar{a}^{|H|m}=\bar1^m=\bar1.
\end{align*}
(Note that this is a computation in $(\Zmod{p})^{\times}$.)

This implies that we have
\[a^{p-1}\equiv 1 \pmod p.\] Multiplying by $a$, we obtain
\[a^{p}\equiv a\pmod p,\] and hence Fermat’s Little Theorem is proved.

The post Use Lagrange's Theorem to Prove Fermat's Little Theorem first appeared on Problems in Mathematics.

Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.

Proof.

Since $G$ acts on $X$, it induces a permutation representation
\[\rho: G \to S_{X}.\]

Let $N=\ker(\rho)$ be the kernel of $\rho$.
Since a kernel is normal in $G$ and $G$ is simple, we have either $N=\{e\}$ or $N=G$.

If $N=G$, then for any $g\in G$ we have $\rho(g)$ is a trivial action, that is, $g\cdot x=x$ for any $X$.
This contradicts the assumption that $G$ acts nontrivially on $X$.
Hence we have $N=\{e\}$, and it follows that the homomorphism $\rho$ is injective.

Thus we have
\[G \cong \mathrm{im} (\rho) < S_{X}.\] Since $S_{X}$ is a finite group and $G$ is isomorphic to its subgroup, the group $G$ is finite.
By Lagrange’s theorem, the order $|G|=|\mathrm{im}(\rho)|$ of $G$ divides the order $|S_{X}|=|X|!$ of $S_{X}$.

The post Nontrivial Action of a Simple Group on a Finite Set first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.

Then prove that any subgroup of index $p$ is a normal subgroup of $G$.
 

Hint.

Consider the action of the group $G$ on the left cosets $G/H$ by left multiplication.

Proof.

Let $H$ be a subgroup of index $p$.
Then the group $G$ acts on the left cosets $G/H$ by left multiplication.

It induces the permutation representation $\rho: G \to S_p$.

Let $K=\ker \rho$ be the kernel of $\rho$.
Since $kH=H$ for $k\in K$, we have $K\subset H$.
Let $[H:K]=m$.

By the first isomorphism theorem, the quotient group $G/K$ is isomorphic to the subgroup of $S_p$, thus $[G:K]$ divides $|S_p|=p!$ by Lagrange’s theorem.
Since $[G:K]=[G:H][H:K]=pm$, we have $pm|p!$ and hence $m|(p-1)!$.

If $m$ has a prime factor $q$, then $q\geq p$ since the minimality of $p$ but the factors of $(p-1)!$ are only prime numbers less than $p$.
Thus $m|(p-1)!$ implies that $m=1$, hence $H=K$. Therefore $H$ is normal since a kernel is always normal.

The post A Subgroup of the Smallest Prime Divisor Index of a Group is Normal first appeared on Problems in Mathematics.

Let $G$ be a group of order $|G|=pq$, where $p$ and $q$ are (not necessarily distinct) prime numbers.

Then show that $G$ is either abelian group or the center $Z(G)=1$.

Hint.

Use the result of the problem “If the Quotient by the Center is Cyclic, then the Group is Abelian”.

Proof.

Since the center $Z(G)$ is a (normal) subgroup  of $G$, the order of $Z(G)$ divides the order of $G$ by Lagrange’s theorem.
Thus the order of $Z(G)$ is one of $1,p,q,pq$.

Suppose that $Z(G)\neq 1$.
Then the order of the quotient group $G/Z(G)$ is one of $1,p,q$.
Hence the group $G/Z(G)$ is a cyclic group.

We conclude that $G$ is abelian group by Problem “If the Quotient by the Center is Cyclic, then the Group is Abelian”.

Therefore, either $Z(G)=1$ or $G$ is abelian.

The post Group of Order $pq$ is Either Abelian or the Center is Trivial first appeared on Problems in Mathematics.