# Group does not Necessarily have Subgroup of Order of Divisor of its Order/Proof 2

## 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

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 $6$.

$\blacksquare$