Cauchy's theorem: if a prime $p$ divides $\mathrm{\mid }G\mathrm{\mid }$, then $G$ has an element of order $p$

If a finite $p$-group $P$ acts on a finite set $X$, then $\mathrm{\mid }X\mathrm{\mid }\equiv \mathrm{\mid }{X}^P\mathrm{\mid }(\mathrm{m}\mathrm{o}\mathrm{d}p)$

A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$

For every natural $n$, $(\mathbb{Z}\mathrm{/}n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold

The congruence class $[a{]}_n$ and the quotient set $\mathbb{Z}\mathrm{/}n$

For $n\ge 1$, every class in $\mathbb{Z}\mathrm{/}n$ has one representative $r$ with $0\le r<n$, so $\mid \mathbb{Z}\mathrm{/}n\mid =n$; while $\mathbb{Z}\mathrm{/}0$ is in bijection with $\mathbb{Z}$

The set $A^B$ of functions $B\to A$ between finite sets is finite, with $\mid A^B\mid =\mid A{\mid }^{\mid B\mid }$

Exponentiation of natural numbers, $m^n$, and its agreement with the integer power in $\mathbb{R}$

Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$

Congruence modulo an integer: $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$ when $n\mid (a-b)$, including the moduli $0$ and $1$

Divisibility in $\mathbb{Z}$: $d\mid a$ when $a=dq$ for some integer $q$

The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ when no positive power of $g$ is the identity

If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$