The Order of a Fin. Group is the Prod. of the Ord. of its Comp. Factors

Math Online's 3000th Page - 5:30 PM CST on August 6th, 2018

The Order of a Finite Group is the Product of the Orders of its Composition Factors

Theorem 1: Let $G$ be a finite group and let $\{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G$ be a composition series of $G$. Then $\displaystyle{|G| = \prod_{i=0}^{n-1} |G_{i+1}/G_i|}$.

On The Jordan-Hölder Theorem page we proved that every finite group has a composition series.

  • Proof: We carry this proof out by induction on the length of the composition series.
  • Let $P(n)$ be the statement that if $G$ is finite group and has a composition series of length $n$, then $|G|$ is the product of the orders of the composition factors of that composition series.
  • Base Step: Consider $P(1)$. Let $G$ be a group with a composition series of length $1$, say $\{ e \} = G_0 \leq G_1 = G$. Then $G_1/G_0 = G/\{ e \} \cong G$. So indeed:
\begin{align} \quad |G| = \prod_{i=0}^{0} |G_{i+1}/G_i| = |G_1/G_0| = |G/\{e\}| = |G| \end{align}
  • Suppose that for some $n > 1$ we have that $P(1)$, $P(2)$, …, $P(n-1)$ is true. Let $G$ be a finite group with a composition series of length $n$, say:
\begin{align} \quad \{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G \end{align}
  • Without loss of generality we can assume that $G_{i+1} \neq G_i$ for all $0 \leq i \leq i+1$, since if $G_{i+1} = G_i$ then $|G_{i+1}/G_i| = 1$. So if $G_{i+1} \neq G_i$ for all $0 \leq i \leq i+1$ then notice that $G_{n-1}$ has composition series:
\begin{align} \quad \{ e \} = G_0 \leq G_1 \leq ... \leq G_{n-1} \end{align}
  • Since $G_{n-1} \leq G_n$ we have that $G_{n-1}$ is a finite group with the above composition series, and so by the induction hypothesis we have that:
\begin{align} \quad |G_{n-1}| = \prod_{i=0}^{n-2} |G_{i+1}/G_i| \end{align}
  • But then by Lagrange's Theorem, since $G_{n-1}$ is a subgroup of $G_n = G$ we have that:
\begin{align} \quad |G| = |G_{n-1}|[G:G_{n-1}] = \prod_{i=0}^{n-2}|G_{i+1}/G_i| \cdot \frac{|G_n|}{|G_{n-1}|} = \prod_{i=0}^{n-2} |G_n/G_{n-1}| = \prod_{i=0}^{n-1} |G_{i+1}/G_i| \end{align}
  • So $P(n)$ is true.
  • Conclusion: By the principle of mathematical induction we have that $P(n)$ is true for all $n \in \mathbb{N}$. $\blacksquare$
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