# simplicity of the alternating groups

###### Theorem 1.

If $n\geq 5$, then the alternating group  on $n$ symbols, $A_{n}$, is simple.

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

 $\pi=(a_{1},a_{2},\ldots,a_{k})(b_{1},b_{2},\ldots,b_{l})\ldots(c_{1},\ldots,c_% {m})$

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

 $\sigma\pi\sigma^{-1}=(\sigma(a_{1}),\sigma(a_{2}),\ldots,\sigma(a_{k}))\,(% \sigma(b_{1}),\sigma(b_{2}),\ldots\sigma(b_{l}))\,\ldots(\sigma(c_{1}),\ldots,% \sigma(c_{m}))$

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 $3$ 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  $N\triangleleft A_{n}$ contains a $3$-cycle, then $N=A_{n}$.

###### 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 $n\geq 5$) 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 $N\triangleleft A_{n}$. Hence, by previous lemma $N=A_{n}$ as required. ∎

## Proof of theorem.

Let $N\triangleleft A_{n}$ 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

 $\pi=(a_{1},a_{2},a_{3},a_{4},\ldots)\ldots$

Upon conjugation by $(a_{1},a_{2},a_{3})\in A_{n}$, we obtain

 $\pi^{\prime}=(a_{1},a_{2},a_{3})\pi(a_{3},a_{2},a_{1})=(a_{2},a_{3},a_{1},a_{4% },\ldots)\ldots$

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

 $\pi=(a,b,c)(d,e,f)\ldots$

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

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

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

## Case 3.

Suppose that there exists a $\pi\in N$ whose disjoint cycle decomposition consists of exactly one $3$-cycle and an even (possibly zero) number of transpositions. Hence, $\pi\pi$ is a $3$-cycle. Lemma 3 can then be applied to complete      the proof.

## 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 $n\geq 5$) 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

 $\pi=(a_{1},b_{1})(a_{2},b_{2})(a_{3},b_{3})(a_{4},b_{4})\ldots$

This time we conjugate by $(a_{2},b_{1})(a_{3},b_{2})$.

 $\pi^{\prime}=(a_{2},b_{1})(a_{3},b_{2})\pi(a_{3},b_{2})(a_{2},b_{1})=(a_{1},a_% {2})(a_{3},b_{1})(b_{2},b_{3})(a_{4},b_{4})\ldots.$

Observe that

 $\pi^{\prime}\pi=(a_{1},a_{3},b_{2})(a_{2},b_{3},b_{1})\ldots,$

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 SimplicityOfTheAlternatingGroups 2013-03-22 13:07:57 2013-03-22 13:07:57 rmilson (146) rmilson (146) 16 rmilson (146) Result msc 20D06 msc 20E32 ExamplesOfFiniteSimpleGroups