Group has Subgroups of All Prime Power Factors

Theorem

Let $p$ be a prime.

Let $G$ be a finite group of order $n$.

If $p^k \divides n$ then $G$ has at least one subgroup of order $p^k$.

Proof

From Composition Series of Group of Prime Power Order, a $p$-group has subgroups corresponding to every divisor of its order.

Thus, taken with the First Sylow Theorem, a finite group has a subgroup corresponding to every prime power divisor of its order.

$\blacksquare$

Sources

• 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.6$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Lagrange's theorem: 2.
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Lagrange's theorem: 2.