Lagrange's theorem and generatorsAQA A-Level Further Maths: Revision notes

Lagrange's theorem: if $H$ is a subgroup of a finite group $G$, then the order of $H$ divides the order of $G$, that is $|H|$ divides $|G|$. Use it to rule out subgroups. A group of order $12$ can have subgroups of order $1,2,3,4,6,12$ but never of order $5$, $8$ or $9$. Lagrange does not guarantee that a subgroup exists for each divisor, but it does say which orders are impossible. If $K$ contains a subgroup $H$ then $H$ is a subgroup of $K$, so $|H|$ divides $|K|$ and $|K|$ divides $|G|$. For example, if $G$ has order $12$ and $K$ contains a subgroup of order $6$, then $|K|$ is $6$ or $12$.

The order of an element $a$ equals the order of the subgroup $\langle a\rangle=\{e,a,a^2,\dots\}$ that it generates. So by Lagrange, the order of every element divides $|G|$. In a group of order $12$, no element has order $5$ or $8$. If $|G|=p$, a prime, the only divisors are $1$ and $p$, so the only subgroups are $\{e\}$ and $G$. Any non-identity element has order $p$, so it generates $G$: a group of prime order is cyclic. A group of order $7$ has $6$ elements of order $7$ and one of order $1$.

An element $g$ is a generator of $G$ if $\langle g\rangle=G$, that is, if every element is a power of $g$. Equivalently, $g$ has order $|G|$. A group with a generator is cyclic. To show $g$ generates $G$, list its powers and show you reach every element, or show that its order equals $|G|$. Example: in $\{1,2,\dots,10\}$ under multiplication modulo $11$, the powers of $2$ are $2,4,8,5,10,9,7,3,6,1$: all ten elements, so $2$ is a generator and the group is cyclic.

If $G=\langle a\rangle$ has order $n$, then $a^k$ has order $\frac{n}{\gcd(k,n)}$. So $a^k$ is a generator exactly when $\gcd(k,n)=1$. For $n=18$ the generators are $a,a^5,a^7,a^{11},a^{13},a^{17}$. Once one generator $g$ is known, the others are the powers $g^k$ with $\gcd(k,n)=1$. In the modulo $11$ group, $2$ generates, so the generators are $2^1,2^3,2^7,2^9=2,8,7,6$. The subgroup $\langle a^k\rangle$ has order $\frac{n}{\gcd(k,n)}$. In a cyclic group of order $18$, $\langle a^3\rangle=\{e,a^3,a^6,a^9,a^{12},a^{15}\}$ has order $6$, and there is exactly one subgroup for each divisor of $18$.

$G=\{1,\dots,10\}$ under multiplication modulo $11$ has order $10$, generated by $2$. By Lagrange, subgroup orders are $1,2,5,10$. Order $5$: $2^2=4$ has order $5$, and its powers $4,5,9,3,1$ give $\{1,3,4,5,9\}$. Order $2$: $2^5=10$ has order $2$, so the subgroup is $\{1,10\}$. It is the only one, because $x^2\equiv1\pmod{11}$ gives $(x-1)(x+1)\equiv0$, so $x\equiv1$ or $10$, since $11$ is prime. These are the proper non-trivial subgroups of $G$.