# The alternating group on five elements is simple

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 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 $$5$$.

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Parents:

• Alternating group

The alternating group is the only normal subgroup of the symmetric group (on five or more generators).

• This is a test comment

• 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?

• Most technical version goes onto the primary page (this one). Easier versions get their own lenses. You could title the lens “Easy proof” or “Simple proof’.