Corollaries to Lagrange's Theorem

# Corollaries to Lagrange's Theorem

Recall from the Lagrange's Theorem page that if $(G, \cdot)$ is a finite group and $(H, \cdot)$ is a subgroup then the number of elements in $H$ must divide the number of elements in $G$ and moreover:

(1)
\begin{align} \quad \mid G \mid = [G : H] \mid H \mid \end{align}

We will now prove some amazing corollaries relating to Lagrange's theorem.

 Corollary 1: Let $(G, \cdot)$ be a finite group and let $(H, \cdot)$ be a subgroup. Then the index of $H$ is $G$ is given by $\displaystyle{[G : H] = \frac{\mid G \mid}{\mid H \mid}}$.
• Proof: Rearrange the formula in the proof of Lagrange's theorem. $\blacksquare$
 Corollary 2: Let $(G, \cdot)$ be a finite group and let $g \in G$. Then the order of the element $g$ must divide $\mid G \mid$.

The order of an element $g$ is the smallest positive integer $n$ such that $g^n = e$ (where of course, $e$ is the identity element for the group.

• Proof: The order of the element $g$ is the smallest positive integer $n$ such that $g^n = e$. The generated subgroup $(<g>, \cdot)$ is such that $\mid <g> \mid = n$. So by Lagrange's theorem $n = \mid < g > \mid$ must divide $\mid G \mid$, i.e., the order of $g$ must divide $\mid G \mid$.
 Corollary 3: Let $(G, \cdot)$ be a finite group of order $\mid G \mid = p$ where $p$ is a prime number. Then $G$ is a cyclic group.
• Proof: Let $g \in G$ be any non-identity element in $G$ (which exists since $p \geq 2$). By Lagrange's Theorem, the subgroup $(<g>, \cdot)$ must be such that $\mid <g> \mid$ divides $\mid G \mid = p$. But the only positive divisors of $p$ are $1$ and $p$.
• So if $\mid <g> \mid = 1$ then $g = e$ which is a contradiction since we assumed that $g$ is not the identity for the group. So $\mid <g> \mid = p$, i.e., $G = <g>$. So $G$ is a cyclic group.
 Corollary 4: Let $(G, \cdot)$ be a finite group and let $(H, \cdot)$ and $(I, \cdot)$ be subgroups of $G$ such that $I \subseteq H \subseteq G$. Then $[G : I] = [G : H][H : I]$.
• Proof: By Lagrange's theorem we have that:
(2)
\begin{align} \quad [G:I] &= \frac{\mid G \mid}{\mid I \mid} = \frac{\mid G \mid}{\mid H \mid} \cdot \frac{\mid H \mid}{\mid I \mid} = [G : H] [H : I] \quad \blacksquare \end{align}