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.

Normal Subgroup Whose Order is Relatively Prime to Its Index
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 […]

Group of Order $pq$ is Either Abelian or the Center is Trivial
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 […]

The Order of $ab$ and $ba$ in a Group are the Same
Let $G$ be a finite group. Let $a, b$ be elements of $G$.
Prove that the order of $ab$ is equal to the order of $ba$.
(Of course do not assume that $G$ is an abelian group.)
Proof.
Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,
\[(ab)^n=e, […]

Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57
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 […]

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

Nontrivial Action of a Simple Group on a Finite Set
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 […]

If the Order of a Group is Even, then the Number of Elements of Order 2 is Odd
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 […]