# Group does not Necessarily have Subgroup of Order of Divisor of its Order

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## 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 1

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.

$\blacksquare$

## Proof 2

Consider the symmetric group $S_4$.

Then the order of the alternating group $A_4$ is $12$.

We list the subgroups of $A_4$:

The subsets of $A_4$ which form subgroups of $A_4$ are as follows:

Trivial:

\(\ds \) | \(\) | \(\ds \set e\) | Trivial Subgroup is Subgroup | |||||||||||

\(\ds \) | \(\) | \(\ds A_4\) | Group is Subgroup of Itself |

\(\ds \) | \(\) | \(\ds \set {e, t}\) | as $t^2 = e$ | |||||||||||

\(\ds \) | \(\) | \(\ds \set {e, u}\) | as $u^2 = e$ | |||||||||||

\(\ds \) | \(\) | \(\ds \set {e, v}\) | as $v^2 = e$ |

\(\ds \) | \(\) | \(\ds \set {e, a, p}\) | as $a^2 = p$, $a^3 = a p = e$ | |||||||||||

\(\ds \) | \(\) | \(\ds \set {e, b, s}\) | as $b^2 = s$, $b^3 = b s = e$ | |||||||||||

\(\ds \) | \(\) | \(\ds \set {e, c, q}\) | as $c^2 = q$, $c^3 = c q = e$ | |||||||||||

\(\ds \) | \(\) | \(\ds \set {e, d, r}\) | as $d^2 = r$, $d^3 = d r = e$ |

\(\ds \) | \(\) | \(\ds \set {e, t, u, v}\) | Klein $4$-Group |

Now $6$ divides $12$.

But there is no subgroup of $A_4$ of order $6$.

$\blacksquare$

## Also see

## Sources

- 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.5$ - 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.**