Alternating Group is Simple except on 4 Letters

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Theorem

Let $n$ be an integer such that $n \ne 4$.


Then the $n$th alternating group $A_n$ is simple.


Proof

Recall that a group is simple if its normal subgroups are itself and the trivial subgroup.

Let $n < 4$.

$A_2$ is the trivial group and hence simple.

$A_3$ is the cyclic group of order $3$, hence a prime group.

By Prime Group is Simple, $A_3$ is simple.


We note that $A_4$ is a special case.

From Normality of Subgroups of Alternating Group on 4 Letters, $A_4$ has a proper normal subgroup $K_4$, and so is not simple.

$\Box$


It remains to investigate the case where $n \ge 5$.

Let $N$ be a non-trivial normal subgroup of $A_n$.

It is to be shown that $N = A_n$.


Lemma 1

Let $\alpha \in A_n$ be a permutation on $\N_n$ such that $\map \alpha 1 = 2$.

Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.

Then the permutation $\beta^{-1} \alpha^{-1} \beta \alpha$ fixes $1$.

$\Box$


Lemma 2

Let $\alpha \in A_n$ be the permutation on $\N_n$ in the form:

$\alpha = \tuple {1, 2} \tuple {3, 4}$

Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.

Then:

$\beta^{-1} \alpha^{-1} \beta \alpha = \beta$

$\Box$


Step $1$: $N$ contains a $3$-cycle

The first step is to show that $N$ contains a $3$-cycle.

Let $e$ denote the identity element of $A_n$.

Let $\alpha \ne e$ be an element of $N$ which fixes as many elements of $\N_n$ as possible.

From Existence and Uniqueness of Cycle Decomposition, we can express $\alpha$ in the form:

$\alpha = \alpha_1 \alpha_2 \dotsm \alpha_s$

where each of the $\alpha_i$ are disjoint cycles.

Let it be assumed that $\alpha_1, \ldots, \alpha_s$ are arranged in order of increasing length.

We can renumber $\N_n$ if we need to, so as to fix it so that:

$\alpha_1 = \tuple {1, 2, \ldots, k}$

and when $s > 1$ that:

$\alpha_2 = \tuple {k + 1, k + 2, \ldots, l}$


We investigate several cases.


Case $1$
$\alpha$ moves each of $1, 2, 3, 4, 5$, and maybe others.

This happens in one of the following situations:

when $s > 2$
when $s = 2$ and $\alpha = \tuple {1, 2, \ldots, k} \tuple {k + 1, k + 2, \ldots, l}$ where $l > 4$
when $s = 1$ and $\alpha = \alpha_1 = \tuple {1, 2, \ldots, k}$ where $k > 4$.

Let $\beta = \tuple {3, 4, 5}$.

We started with the assumption that $N$ is normal.

Hence we have that:

$\beta^{-1} \alpha^{-1} \beta \in N$

Thus:

$\beta^{-1} \alpha^{-1} \beta \alpha \in N$

However, from lemma 1, $\beta^{-1} \alpha^{-1} \beta \alpha$ leaves $1$ fixed.

This is in addition to all the other elements fixed by $\alpha$.

But Case $1$ was specifically stated such that $\alpha$ moves each of $1, 2, 3, 4, 5$.

So Case $1$ cannot happen.


Case $2$
$\alpha$ moves each of $1, 2, 3, 4$ and no others.

This happens only when $\alpha = \tuple {1, 2} \tuple {3, 4}$, because $\tuple {1, 2, 3, 4}$ is an odd permutation.

Again, let $\beta = \tuple {3, 4, 5}$.

We started with the assumption that $N$ is normal.

Hence we have that:

$\beta^{-1} \alpha^{-1} \beta \in N$

Thus:

$\beta^{-1} \alpha^{-1} \beta \alpha \in N$

However, from lemma 2:

$\beta^{-1} \alpha^{-1} \beta \alpha = \beta$

Thus $\beta \in N$ and $\beta$ moves fewer elements of $\N_n$ than $\alpha$.

But $\alpha$ was defined as fixing as many elements of $\N_n$ as possible.

So Case $2$ cannot happen.


Case $3$
$\alpha$ moves each of $1, 2, 3$ and no others.

This happens only when $\alpha = \tuple {1, 2, 3}$.

There can be no other cases now that Case $1$ and Case $2$ have been eliminated.

Thus we have shown that $N$ contains a $3$-cycle.

Thus the first step has been completed.

$\Box$


Step $2$: $N$ contains every $3$-cycle

Let $\sigma$ be an arbitrary even permutation, presented in two-row notation:

$\sigma = \begin {pmatrix} 1 & 2 & 3 & \cdots \\ i & j & k & \cdots \end {pmatrix}$

Then $\sigma \tuple {1, 2, 3} \sigma^{-1} = \tuple {i, j, k}$ is an element of the normal subgroup $N$.

We have that $i$, $j$ and $k$ are arbitrary.

So it follows that every $3$-cycle $\tuple {i, j, k}$ is an element of $N$.

It follows from Even Permutation is Product of 3-Cycles that every element of $A_n$ is in $N$.

That is:

$N = A_n$

and the result follows.

$\blacksquare$


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