Order, subgroups and Lagrange's theoremEdexcel A-Level Further Maths: Revision notes

The order of a group $G$, written $|G|$, is the number of elements in it. The order of an element $g$ is the smallest positive integer $n$ with $g^n=e$ (in additive notation, $ng=0$). If no such $n$ exists the element has infinite order, but in a finite group every element has a finite order. Example: in $\{1,\ldots,10\}$ under multiplication modulo 11, $3^1=3$, $3^2=9$, $3^3=5$, $3^4=4$, $3^5=1$, so $3$ has order $5$. The identity always has order $1$ and is the only element of order $1$.

A subgroup $H$ of $G$ is a subset of $G$ that is itself a group under the same operation. To check, show $H$ is non-empty, closed under the operation, contains the identity and contains the inverse of each of its elements (associativity is inherited from $G$). Every group has the trivial subgroups $\{e\}$ and $G$ itself; any other subgroup is called proper. The powers of one element $g$ form a subgroup $\langle g\rangle=\{e,g,g^2,\ldots\}$, called the cyclic subgroup generated by $g$. Its order equals the order of $g$. Example: in the symmetries of an equilateral triangle, $\{e,r,r^2\}$ is a subgroup of order $3$ and $\{e,m_1\}$ a subgroup of order $2$.

Lagrange's theorem: if $G$ is a finite group and $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$. So a group of order $12$ can only have subgroups of order $1,2,3,4,6,12$. There is no subgroup of order $5$, $7$, $8$, $9$, $10$ or $11$. The theorem tells you which orders are impossible, but not that every divisor occurs.

The cyclic subgroup $\langle g\rangle$ has order equal to the order of $g$, so by Lagrange's theorem the order of every element divides the order of the group. In a group of order $24$ there is no element of order $16$. If an element has order equal to $|G|$, then $\langle g\rangle=G$, so $G$ is cyclic. A group of prime order is cyclic, since every non-identity element has order dividing a prime $p$, so equal to $p$, and so generates the whole group; its only subgroups are the trivial ones.

Find all subgroups of $\{0,1,\ldots,11\}$ under addition modulo 12. By Lagrange the orders are $1,2,3,4,6,12$. The subgroups are $\{0\}$, $\{0,6\}$, $\{0,4,8\}$, $\{0,3,6,9\}$, $\{0,2,4,6,8,10\}$ and $G$. Element orders: $3$ has order $4$ ($3,6,9,0$), $5$ has order $12$ ($5k\equiv0$ first at $k=12$), $8$ has order $3$ ($8,4,0$). Each order divides $12$ as Lagrange predicts, and $5$ generates the whole group.