Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 1

Theorem

Let $G$ be a finite group whose order is $n$.

Let $d$ be a divisor of $n$.

Then it is not necessarily the case that $G$ has a subgroup of order $d$.

Proof

Proof by Counterexample:

Consider $S_5$, the symmetric group on $5$ letters.

By Order of Symmetric Group, $\order {S_5} = 5! = 120$.

We have that $120 = 8 \times 15$ and so $15$ is a divisor of $120$.

However, Symmetric Group on 5 Letters has no Subgroup of Order 15.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44$. Some consequences of Lagrange's Theorem