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: $1.5$