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

# Proof

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 $~$5$~$.

## Comments

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?