Groups formulas

A binary operation on a set of $n$ elements is a filled $n\times n$ table, each of its $n^2$ entries chosen from the $n$ elements. A commutative one is fixed by the $n(n+1)/2$ entries on and above the main diagonal. For $n=2$: $16$ in all, $8$ commutative.

Holds in every group; note the reversed order. If the group is abelian the order does not matter. Also $(a^{-1}{)}^{-1}=a$.

For a finite group $G$ and any subgroup $H$; $[G:H]$ is the number of distinct left cosets of $H$ in $G$.

In every finite group, by Lagrange applied to $⟨a⟩$. Also $a^k=e⟺o(a)\mid k$.

Holds in any group. In ${\mathbb{Z}}_n$ the element $k$ has order $n/\gcd (n,k)$. If $G=⟨a⟩$ has order $n$, the element $a^k$ generates $G$ exactly when $\gcd (n,k)=1$.

For a cyclic group of order $n$, where $\varphi (n)$ is Euler's totient (the count of $k$ with $1\le k\le n$ and $\gcd (k,n)=1$).

The product runs over the distinct primes $p$ dividing $n$. For a prime $p$, $\varphi (p)=p-1$ and $\varphi (p^k)=p^k-p^{k-1}$.

A cyclic group of order $n$ has exactly one subgroup for each positive divisor $d$ of $n$, so the number of subgroups equals the number of divisors of $n$. Every subgroup of a cyclic group is cyclic, and the group has exactly $\varphi (d)$ elements of order $d$ for each $d\mid n$.

In $G\times H$ with finite $o(a)$ and $o(b)$. Consequently ${\mathbb{Z}}_m\times {\mathbb{Z}}_n$ is cyclic if and only if $\gcd (m,n)=1$.

Any non-identity element has order dividing $p$, so order $p$, hence generates $G$. Every non-identity element is a generator, giving $p-1$ generators.