The alternating group on five elements is simple

by Patrick Stevens Jun 17 2016 updated Jun 28 2016

The smallest (nontrivial) simple group is the alternating group on five elements.

The Alternating group $~$A_5$~$ on five elements is simple.


Recall that $~$A_5$~$ has order $~$60$~$, so Lagrange's theorem states that any subgroup of $~$A_5$~$ has order dividing $~$60$~$.

Suppose $~$H$~$ is a normal subgroup of $~$A_5$~$, which is not the trivial subgroup $~$\{ e \}$~$. If $~$H$~$ has order divisible by $~$3$~$, then by Cauchy's theorem there is a $~$3$~$-cycle in $~$H$~$ (because the $~$3$~$-cycles are the only elements with order $~$3$~$ in $~$A_5$~$). Because $~$H$~$ is a union of conjugacy classes, and because the $~$3$~$-cycles [4kv form a conjugacy class in $~$A_n$~$ for $~$n > 4$~$], $~$H$~$ would therefore contain every $~$3$~$-cycle; but then it would be the entire alternating group.

If instead $~$H$~$ has order divisible by $~$2$~$, then there is a double transposition such as $~$(12)(34)$~$ in $~$H$~$, since these are the only elements of order $~$2$~$ in $~$A_5$~$. But then $~$H$~$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $~$(12)(34)$~$ and $~$(15)(34)$~$, so it contains $~$(15)(34)(12)(34) = (125)$~$. Hence as before $~$H$~$ contains every $~$3$~$-cycle so is the entire alternating group.

So $~$H$~$ must have order exactly $~$5$~$, by Lagrange's theorem; so it contains an element of order $~$5$~$ since prime order groups are cyclic.

The only such elements of $~$A_n$~$ are $~$5$~$-cycles; but the conjugacy class of a $~$5$~$-cycle is of size $~$12$~$, which is too big to fit in $~$H$~$ which has size $~$5$~$.


Patrick Stevens

This is definitely a page which admits two lenses: the "easy" proof and the "theory-heavy" proof. What kind of lens design might people use?