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

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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

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.


Proof 2

Proof by Counterexample:

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:


\(\ds \) \(\) \(\ds \set e\) Trivial Subgroup is Subgroup
\(\ds \) \(\) \(\ds A_4\) Group is Subgroup of Itself

Order $2$ subgroups:

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

Order $3$ subgroups:

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

Order $4$ subgroup:

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

Now $6$ divides $12$.

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


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