# simplicity of the alternating groups

###### Theorem 1.

If $n\mathrm{\ge}\mathrm{5}$, then the alternating group^{}
on $n$ symbols, ${A}_{n}$, is simple.

Throughout the proof we extensively employ the cycle notation, with
composition^{} on the left, as is usual. The symmetric group^{} on $n$
symbols is denoted by ${S}_{n}$.

The following observation will be useful. Let $\pi $ be a permutation^{}
written as disjoint cycles

$$\pi =({a}_{1},{a}_{2},\mathrm{\dots},{a}_{k})({b}_{1},{b}_{2},\mathrm{\dots},{b}_{l})\mathrm{\dots}({c}_{1},\mathrm{\dots},{c}_{m})$$ |

It is easy to check that for every permutation $\sigma \in {S}_{n}$ we have

$$\sigma \pi {\sigma}^{-1}=(\sigma ({a}_{1}),\sigma ({a}_{2}),\mathrm{\dots},\sigma ({a}_{k}))(\sigma ({b}_{1}),\sigma ({b}_{2}),\mathrm{\dots}\sigma ({b}_{l}))\mathrm{\dots}(\sigma ({c}_{1}),\mathrm{\dots},\sigma ({c}_{m}))$$ |

As a consequence, two permutations of ${S}_{n}$ are conjugate^{} exactly when
they have the same cycle type.

Two preliminary results will also be necessary.

###### Lemma 2.

The set of cycles of length $\mathrm{3}$ generates ${A}_{n}$.

###### Proof.

A product^{} of $3$-cycles is an even permutation^{}, so the subgroup^{} generated by all $3$-cycles is therefore contained in ${A}_{n}$. For the reverse inclusion, by definition every even permutation is the product of even number^{} of transpositions^{}. Thus, it suffices to show that the product of two transpositions can be written as a product of $3$-cycles. There are two possibilities. Either the two transpositions move an element in common, say $(a,b)$ and $(a,c)$, or the two transpositions are disjoint, say $(a,b)$ and $(c,d)$. In the former case,

$$(a,b)(a,c)=(a,c,b),$$ |

and in the latter,

$$(a,b)(c,d)=(a,b,d)(c,b,d).$$ |

This establishes the first lemma. ∎

###### Lemma 3.

If a normal subgroup^{} $N\mathrm{\u25c1}{A}_{n}$ contains a $\mathrm{3}$-cycle, then
$N\mathrm{=}{A}_{n}$.

###### Proof.

We will show that if $(a,b,c)\in N$, then the assumption^{} of normality implies that any other $({a}^{\prime},{b}^{\prime},{c}^{\prime})\in N$. This is easy to show, because there is some permutation in $\sigma \in {S}_{n}$ that under conjugation^{} takes $(a,b,c)$ to $({a}^{\prime},{b}^{\prime},{c}^{\prime})$, that is

$$\sigma (a,b,c){\sigma}^{-1}=(\sigma (a),\sigma (b),\sigma (c))=({a}^{\prime},{b}^{\prime},{c}^{\prime}).$$ |

In case $\sigma $ is odd, then (because $n\ge 5$) we can choose some transposition $(d,e)\in {A}_{n}$ disjoint from $({a}^{\prime},{b}^{\prime},{c}^{\prime})$ so that

$$\sigma (a,b,c){\sigma}^{-1}=(d,e)({a}^{\prime},{b}^{\prime},{c}^{\prime})(d,e),$$ |

that is,

$${\sigma}^{\prime}(a,b,c){\sigma}^{\prime -1}=(d,e)\sigma (a,b,c){\sigma}^{-1}(d,e)=({a}^{\prime},{b}^{\prime},{c}^{\prime})$$ |

where ${\sigma}^{\prime}$ is even. This means that $N$ contains all $3$-cycles, as $N\u25c1{A}_{n}$. Hence, by previous lemma $N={A}_{n}$ as required. ∎

## Proof of theorem.

Let $N\u25c1{A}_{n}$ be a non-trivial normal subgroup. We will show that $N={A}_{n}$. The proof now proceeds by cases. In each case, the normality of $N$ will allow us to reduce the proof to Lemma 2 or to one of the previous cases.

## Case 1.

Suppose that there exists a $\pi \in N$ that, when written as disjoint cycles, has a cycle of length at least $4$, say

$$\pi =({a}_{1},{a}_{2},{a}_{3},{a}_{4},\mathrm{\dots})\mathrm{\dots}$$ |

Upon conjugation by $({a}_{1},{a}_{2},{a}_{3})\in {A}_{n}$, we obtain

$${\pi}^{\prime}=({a}_{1},{a}_{2},{a}_{3})\pi ({a}_{3},{a}_{2},{a}_{1})=({a}_{2},{a}_{3},{a}_{1},{a}_{4},\mathrm{\dots})\mathrm{\dots}$$ |

Hence, ${\pi}^{\prime}\in N$, and hence ${\pi}^{\prime}{\pi}^{-1}=({a}_{1},{a}_{2},{a}_{4})\in N$ also. Notice that the rest of the cycles cancel. By Lemma 3, $N={A}_{n}$.

## Case 2.

Suppose that there exists a $\pi \in N$ whose disjoint cycle decomposition has at least two cycles of length 3, say

$$\pi =(a,b,c)(d,e,f)\mathrm{\dots}$$ |

Conjugation by $(c,d,e)\in {A}_{n}$ implies that $N$ also contains

$${\pi}^{\prime}=(c,d,e)\pi (e,d,c)=(a,b,d)(e,c,f)\mathrm{\dots}$$ |

Hence, $N$ also contains ${\pi}^{\prime}\pi =(a,d,c,b,f)\mathrm{\dots}$. This reduces the proof to Case 1.

## Case 3.

## Case 4.

Suppose there exists a $\pi \in N$ of the form $\pi =(a,b)(c,d)$. Conjugating by $(a,e,b)$ with $e$ distinct from $a,b,c,d$ (at least one such $e$ exists, as $n\ge 5$) yields

$${\pi}^{\prime}=(a,e,b)\pi (b,e,a)=(a,e)(c,d)\in N.$$ |

Hence ${\pi}^{\prime}\pi =(a,b,e)\in N$. Again, Lemma 3 applies.

## Case 5.

Suppose that $N$ contains a permutation of the form

$$\pi =({a}_{1},{b}_{1})({a}_{2},{b}_{2})({a}_{3},{b}_{3})({a}_{4},{b}_{4})\mathrm{\dots}$$ |

This time we conjugate by $({a}_{2},{b}_{1})({a}_{3},{b}_{2})$.

$${\pi}^{\prime}=({a}_{2},{b}_{1})({a}_{3},{b}_{2})\pi ({a}_{3},{b}_{2})({a}_{2},{b}_{1})=({a}_{1},{a}_{2})({a}_{3},{b}_{1})({b}_{2},{b}_{3})({a}_{4},{b}_{4})\mathrm{\dots}.$$ |

Observe that

$${\pi}^{\prime}\pi =({a}_{1},{a}_{3},{b}_{2})({a}_{2},{b}_{3},{b}_{1})\mathrm{\dots},$$ |

which reduces the proof to Case 2.

Since there exists at least one non-identity $\pi \in N$, and since this $\pi $ is covered by one of the above cases, we conclude that $N={A}_{n}$, as was to be shown.

Title | simplicity of the alternating groups |
---|---|

Canonical name | SimplicityOfTheAlternatingGroups |

Date of creation | 2013-03-22 13:07:57 |

Last modified on | 2013-03-22 13:07:57 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 16 |

Author | rmilson (146) |

Entry type | Result |

Classification | msc 20D06 |

Classification | msc 20E32 |

Related topic | ExamplesOfFiniteSimpleGroups |