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 […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

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 […]