Lagrange's theorem is about subgroups, but its sharpest consequences are about single elements, because every element $a$ comes with its own subgroup $\langle a \rangle$, of order $|a|$.
Why it works
Corollary 2. $a^{|G|} = e$ for every $a \in G$. Proof: $|G| = |a|\,m$ for some $m$, so $a^{|G|} = (a^{|a|})^{m} = e$.
Corollary 3. A group of prime order $p$ is cyclic, and every element other than $e$ generates it. Proof: if $a \ne e$, then $|a|$ divides $p$ and is not $1$, so $|a| = p$ and $\langle a \rangle$ is all of $G$. $\square$
Fermat and Euler, again. Apply Corollary 2 to the group $U(n)$, of order $\varphi(n)$: for $\gcd(a, n) = 1$, $$a^{\varphi(n)} \equiv 1 \pmod{n}.$$ This is Euler's theorem, and with $n = p$ prime it is Fermat's little theorem. The group-theory proof replaces a clever rearrangement argument by one line.
Worked example: A power mod 14
- $3 \in U(14)$, and $|U(14)| = \varphi(14) = 6$, so $3^{6} \equiv 1$.
- $100 = 6 \cdot 16 + 4$, so $3^{100} = (3^{6})^{16}\,3^{4} \equiv 3^{4} = 81$.
- $81 = 5 \cdot 14 + 11$.
Answer: $3^{100} \equiv 11 \pmod{14}$.
Worked example: Element orders in a group of order 35
- They divide $35 = 5 \cdot 7$: $1, 5, 7, 35$.
- Only $e$ has order $1$. If some element has order $35$, the group is cyclic.
Answer: $1$, $5$, $7$ or $35$.
Worked example: Groups of prime order
- Subgroup orders divide $13$: only $1$ and $13$.
- So the subgroups are $\{e\}$ and $G$, and $G \cong \mathbb{Z}_{13}$ is cyclic, generated by any of its $12$ elements other than $e$.
Answer: Two.
These corollaries are why prime order is so restrictive: up to renaming, there is exactly one group of each prime order, the cyclic group. They also sort small groups quickly. In a group of order $4$, element orders divide $4$: either some element has order $4$ and the group is cyclic, or every element other than $e$ has order $2$. So there are just two kinds of group of order $4$, like $\mathbb{Z}_{4}$ and like $U(8)$.
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Key idea