Alternating Group has no Subgroup of Order 6

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $A_4$ denote the alternating group on $4$ letters.

$A_4$ has no subgroup of order $6$.


Proof

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


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


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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Alternating_Group_has_no_Subgroup_of_Order_6&oldid=520521"