# Alternating Group is Simple except on 4 Letters

## 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:

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

$\Box$

### Lemma 3

Let $\rho \in S_n$ be an arbitrary $3$-cycle.

Let $i, j, k \in$ $\N_n$ be such that $\rho = (i, j, k)$.

Then there exists an even permutation $\sigma \in A_n$ such that $\sigma(1) = i$, $\sigma(2) = j$ and $\sigma(3) = k$.

$\Box$

### 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 $\rho \in S_n$ be an arbitrary $3$-cycle.

Let $i, j, k \in$ $\mathbb{N}_n$ be such that $\rho = (i, j, k)$.

By **lemma 3**, there exists an even permutation $\sigma \in A_n$ with the following form when presented in two-row notation:

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

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

We have that $\rho$ is arbitrary.

So it follows that every $3$-cycle 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.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 83$ - 1989: Ephraim J. Borowski and Jonathan M. Borwein:
*Dictionary of Mathematics*... (previous) ... (next): Entry:**alternating group** - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $9$: Permutations: Definition $9.19$: Remark (in passing) - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**alternating group**