PS
Let $n$ be an integer $\geq 5$.
To show that the alternating group $A_{n}$ is simple, we must show that any normal subgroup of $A_{n}$ which contains a nonidentity element $x$ must be all of $A_{n}$. What this comes down to showing is that starting with any such $x$, we can, by using the group operations and the operations of conjugating by various elements of $A_{n}$, eventually come up with every other element $y \in A_{n}$. Now to get from any nonidentity element $x$ to any other element $y$ is a formidable task. To make it more manageable, we shall go via an easy-to-handle intermediate class of group elements. The even permutations that move the smallest number of elements are the cycles of length 3; and the main steps of our proof will in fact be:
Step I. If a normal subgroup $N$ of $A_{n}$ contains a nonidentity element $x$, then it contains a cycle $\left(a_{1}, a_{2}, a_{3}\right)$ of length 3 .
Step II. If a normal subgroup $N$ of $A_{n}$ contains a cycle of length 3 , then it contains every cycle of length $3 .$
Step III. If a subgroup $N$ of $A_{n}$ contains every cycle of length 3, then it is all of $A_{n}$.
Of these, Step I is the most work. The idea is as follows. Let us say that an element $i \in\{1, \ldots, n\}$ is "moved by $\sigma$ " if $\sigma(i) \neq i$; in the contrary case we will say that $i$ is "fixed by $\sigma^{\prime \prime}$; and let us think of a permutation as "small" if it moves few elements of $\{1, \ldots, n\}$. If $x$ is any element of $N$ and $\sigma$ a "small" even permutation, then the conjugate $\sigma \times \sigma^{-1}$, which by normality of $N$ also lies in $N$, will differ only "slightly" from x; i.e., will agree with x except on a small number of elements. Hence if we "divide" the former element by the latter, the resulting permutation, $\left(\sigma x \sigma^{-1}\right) x^{-1}$, will be relatively "small". By applying this principle with a little ingenuity, we will be able to get from an arbitrarily "large" permutation down to a cycle of length $3 .$
Let us begin with a lemma making explicit the computational trick sketched above.
Lemma 1. Let $x, \sigma \in A_{n}$. Then every element $i \in\{1, \ldots, n\}$ that is moved by $\sigma x \sigma^{-1} x^{-1}$ is either an element moved by $\sigma$, or the image under $x$ of an element moved by $\sigma$.
Proof. Suppose $i \in\{1, \ldots, n\}$ is neither an element moved by $\sigma$ nor the image under $x$ of such an element. The latter condition implies that the element whose image under $x$ is $i$, namely $x^{-1}(i)$, is not moved by $\sigma$, hence is not moved by $\sigma^{-1}$ either. Hence we get $\sigma x \sigma^{-1} x^{-1}(i)=\sigma\left(x\left(\sigma^{-1}\left(x^{-1}(i)\right)\right)\right)=$ $\sigma\left(x\left(x^{-1}(i)\right)\right)=\sigma(i)=i$ as required. (The second step, i.e., the second " $=^{\prime \prime}$, uses the fact that $x^{-1}(i)$ is not moved by $\sigma$, the third step uses the identity $\mathrm{xx}^{-1}=\iota$, and the last step, the assumption that $i$ is not moved by $\sigma$.)
Now let $N$ be any nontrivial normal subgroup of $A_{n}$. We shall apply the above lemma in each of a series of cases, to show that in every case, $N$ contains a cycle of length 3 . We shall also use repeatedly the fact that if $x \in N$ and $\sigma \in A_{n}$, then $\sigma x \sigma^{-1} x^{-1}$ is the product of the two elements $\sigma x \sigma^{-1}$ and $x^{-1}$ of $N$, and hence also a member of $N$.
Case A. $N$ contains an element $x$ whose expression as a product of disjoint cycles involves at least one cycle of length greater than $3 .$
Let such a cycle be $\left(a_{1}, a_{2}, a_{3}, a_{4}, \ldots, a_{r}\right)$. (Here $r$ may equal 4 , so the " $\ldots, a_{r}$ " may be empty; but since $r>3, a_{1}, a_{2}, a_{3}, a_{4}$ are distinct.) Thus, $x=\left(a_{1}, a_{2}, a_{3}, a_{4}, \ldots, a_{r}\right) P$, where $P$ is either the identity, or a product of cycles that do not move any of $a_{1}, \ldots, a_{r}$. Let $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$. The elements of $\{1, \ldots, n\}$ that are moved by $\sigma$ are $a_{1}, a_{2}, a_{3}$, and the images under $x$ of those elements are $a_{2}, a_{3}, a_{4}$. Hence by Lemma 1, the permutation $\sigma x \sigma^{-1} x^{-1}$ cannot move any elements but $a_{1}, a_{2}$, $a_{3}, a_{4}$. Therefore we can describe $\sigma x \sigma^{-1} x^{-1}$ by determining what it does on those four elements. Go through this calculation. You will find that $\sigma \times \sigma^{-1} x^{-1}=\left(a_{1}, a_{2}, a_{4}\right)$. Hence in this case $N$ indeed contains a cycle of length 3, as claimed.
Now if a nonidentity element $x \in N$, when expressed as a product of disjoint cycles, does not involve a cycle of length greater than 3 , then it must be a product of disjoint cycles each of length 2 or length 3 . Also note that if an element $x$ is written as a product of disjoint cycles $\sigma_{1} \ldots \sigma_{r}$, then $\sigma_{1}, \ldots, \sigma_{r}$ commute with each other; hence for any integer $d$, we have $x^{d}=\sigma_{1}^{d} \ldots \sigma_{r}^{d}$. In particular, if the $x$ we are interested in here involves both cycles of length 2 and cycles of length 3 , then $x^{2}$ involves only cycles of length 3 (since the square of a cycle of length 2 is the identity, while the square of a cycle of length 3 is another cycle of length 3); and similarly, $x^{3}$ involves only cycles of length 2. Thus by either squaring or cubing such an $x$, we can see that $N$ contains an element which is either a product of one or more disjoint cycles of length 2, or of one or more disjoint cycles of length 3. We now consider those cases:
Case B. $N$ contains an element $x$ which is a product of one or more disjoint cycles of length 3. If $x$ is a single cycle of length 3 , we have what we want. In the contrary case, let us write it as $\left(a_{1}, a_{2}, a_{3}\right)\left(a_{4}, a_{5}, a_{6}\right) P$, where $P$ is again either the identity or a product of cycles that do not move any of $a_{1}, \ldots, a_{6}$. In this case, let us take $\sigma=\left(a_{2}, a_{3}, a_{4}\right)$. Using Lemma 1, we find that $\sigma \times \sigma^{-1} x^{-1}$ can only move some subset of $\left\{a_{1}, \ldots, a_{5}\right\}$. Calculating, we find that $\sigma \times \sigma^{-1}{x}^{-1}=\left(a_{1}, a_{4}, a_{2}, a_{3}, a_{5}\right), a$ cycle of length 5. A pplying the result of Case A, we conclude that $N$ also contains a cycle of length 3 .
Case C. N contains an element $x$ which is a product of one or more disjoint cycles of length 2. In this case there must be at least two such cycles, since a single cycle of length 2 is an odd permutation, while $N$ was assumed a subgroup of the group $A_{n}$ of even permutations. Hence let us write $x=$ $\left(a_{1}, a_{2}\right)\left(a_{3}, a_{4}\right) P$, with $P$ as before. Taking $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$, we find that $\sigma \times \sigma^{-1} x^{-1}=$ $\left(a_{1}, a_{3}\right)\left(a_{2}, a_{4}\right)$.
This is still a product of cycles of length 2, so have we accomplished anything? Y es, we have gotten rid of $P$, and thus have a member of $N$ which moves only 4 elements of $\{1, \ldots, n\}$. Since $n \geq 5$, this means that at least one element is fixed by this permutation. And this puts us in our final case:
Case D. $N$ contains an element $x$ such that the expression for $x$ as a product of disjoint cycles involves a cycle $\left(a_{1}, a_{2}\right)$ of length 2 , and such that $x$ fixes at least one element $a_{3}$. In this situation, let us write $x=\left(a_{1}, a_{2}\right) P$ where $P$ moves none of $a_{1}, a_{2}, a_{3}$, and let $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$. By Lemma 1, $\sigma X \sigma^{-1} x^{-1}$ can move only some subset of $\left\{a_{1}, a_{2}, a_{3}\right\}$. Computing its action on this set, you will find that it equals $\left(a_{1}, a_{3}, a_{2}\right)$, a cycle of length 3 . This completes our proof that in every case, $N$ contains such an element.
For Step II of our proof, we must show that if $N$ contains a cycle $\left(a_{1}, a_{2}, a_{3}\right)$ of length 3 , it contains every cycle $\left(b_{1}, b_{2}, b_{3}\right)$ of that length. To do this, we will need an explicit formula for the result of conjugating a cycle of length 3 by any other permutation $\sigma$. I claim that such a formula is
$$
\sigma\left(a_{1}, a_{2}, a_{3}\right) \sigma^{-1}=\left(\sigma\left(a_{1}\right), \sigma\left(a_{2}\right), \sigma\left(a_{3}\right)\right) \text {. }
$$
Indeed, it is easy to see that the two sides agree on the three elements $\sigma\left(a_{1}\right), \sigma\left(a_{2}\right)$ and $\sigma\left(a_{3}\right)$. If $i$ is not one of these elements, then $\sigma^{-1}(i)$ is not one of $a_{1}, a_{2}$ or $a_{3}$, hence it is not moved by the cycle $\left(a_{1}, a_{2}, a_{3}\right)$, so $\sigma\left(a_{1}, a_{2}, a_{3}\right) \sigma^{-1}(i)=\sigma \sigma^{-1}(i)=i$; so the left-hand side of the above display agrees with the right-hand side on such elements i as well.
Hence if our subgroup $N$ contains a cycle $\left(a_{1}, a_{2}, a_{3}\right)$ and we want to show it to contain some other cycle of length $3,\left(b_{1}, b_{2}, b_{3}\right)$, the obvious thing to do is find a permutation $\sigma$ such that $\sigma\left(a_{1}\right)=b_{1}$, $\sigma\left(a_{2}\right)=b_{2}, \sigma\left(a_{3}\right)=b_{3}$; which we can do, roughly speaking, by writing $\sigma=\left(\begin{array}{lllll}\cdots & a_{1} & \cdots & a_{2} & \cdots & a_{3} & \mathrm{~b}_{2} & \mathrm{~b}_{2} & \mathrm{~b}_{3}\end{array}\right)$ and filling in the remaining entries of the bottom row in any way such that each integer from 1 to $n$ gets used exactly once. (I say "roughly speaking" because the way I have written this permutation assumes $a_{1}<a_{2}<a_{3}$. When that is not so, the columns shown will appear in a different order.)
This will indeed give us a permutation $\sigma$ such that $\sigma\left(a_{1}, a_{2}, a_{3}\right) \sigma^{-1}=\left(b_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}\right)$; and if this $\sigma$ is even, i.e., is a member of $A_{n}$, that will show that the normal subgroup $N$ of $A_{n}$, in addition to $\left(a_{1}, a_{2}, a_{3}\right)$, also contains $\left(b_{1}, b_{2}, b_{3}\right)$. But what if $\sigma$ is odd? There are various ways one can "cure" this; the one we will use is to let $\sigma^{\prime}=\left(b_{2}, b_{3}\right) \sigma$. Since $\left(b_{2}, b_{3}\right)$ is, like $\sigma$, odd, $\sigma^{\prime}$ will be even. It will again send the three elements $a_{1}, a_{2}, a_{3}$ to $b_{1}, b_{2}, b_{3}$, but in a different order; and we see that $\sigma^{\prime}\left(a_{1}, a_{2}, a_{3}\right) \sigma^{\prime-1}=\left(b_{1}, b_{3}, b_{2}\right)$. Now the square of this cycle is the desired cycle $\left(b_{1}, b_{2}, b_{3}\right) ;$ so again, if $N$ contains $\left(a_{1}, a_{2}, a_{3}\right)$, it also contains $\left(b_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}\right)$, completing Step II.
For Step III, we must prove that the cycles of length 3 together generate $A_{n}$. A key fact will be
Lemma 2. If $x$ is a permutation which moves at least three elements, then there exists a cycle $\sigma$ of length 3 such that $\sigma^{-1} x$ moves fewer elements than $x$ does.
Proof. Let $a_{1}$ be an element moved by $x$. If we write $a_{2}=x\left(a_{1}\right)$, this is a second element moved by $x$. Since there are at least three elements moved by $x$, we can choose a third such element, $a_{3}$. Letting $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$, we observe that $\sigma^{-1} x$ does not move any elements that are not moved by $x$ (since neither $x$ nor $\sigma^{-1}$ does); hence if we can prove that $\sigma^{-1} x$ fixes some element that is moved by $x$, we will have the desired conclusion. And indeed, it is immediate from the way we defined $\sigma$ that $\sigma^{-1} x$ fixes $a_{1}$.
Since the only permutations that move fewer than three elements are the identity and the transpositions, we see that every nonidentity element of $A_{n}$ moves at least three elements. Hence if $x$ is a nonidentiy element of $A_{n}$, we can use the above lemma to find a cycle $\sigma_{1}$ of length 3 such that $\sigma_{1}^{-1} x$ moves fewer elements than $x$. If $\sigma_{1}^{-1} x$ is not the identity, we can similarly use the lemma to find a cycle $\sigma_{2}$ of length 3 such that $\sigma_{2}^{-1} \sigma_{1}^{-1} x$ moves fewer elements than $\sigma_{1}^{-1} x$, and so forth. This process must eventually stop, so we must eventually get an expression for the identity permutation as
$$
\sigma_{r}^{-1} \ldots \sigma_{1}^{-1} x
$$
where $\sigma_{1}, \ldots, \sigma_{r}$ are cycles of length 3. This condition can be written $\left(\sigma_{1} \ldots \sigma_{r}\right)^{-1} x=\iota$, equivalently, $x=\sigma_{1} \ldots \sigma_{r}$, showing that the general element $x$ of $A_{n}$ is indeed a product of cycles of length $3 .$
In summary, we have shown that for $n \geq 5$, any normal subgroup $N$ of $A_{n}$ which contains a nonidentity element contains a cycle of length 3, from this that N contains all cycles of length 3, and from this that $N$ contains all elements of $A_{n}$. Thus $A_{n}$ has no proper nontrivial normal subgroup; i.e., it is simple.
Let $n$ be an integer $\geq 5$.
To show that the alternating group $A_{n}$ is simple, we must show that any normal subgroup of $A_{n}$ which contains a nonidentity element $x$ must be all of $A_{n}$. What this comes down to showing is that starting with any such $x$, we can, by using the group operations and the operations of conjugating by various elements of $A_{n}$, eventually come up with every other element $y \in A_{n}$. Now to get from any nonidentity element $x$ to any other element $y$ is a formidable task. To make it more manageable, we shall go via an easy-to-handle intermediate class of group elements. The even permutations that move the smallest number of elements are the cycles of length 3; and the main steps of our proof will in fact be:
Step I. If a normal subgroup $N$ of $A_{n}$ contains a nonidentity element $x$, then it contains a cycle $\left(a_{1}, a_{2}, a_{3}\right)$ of length 3 .
Step II. If a normal subgroup $N$ of $A_{n}$ contains a cycle of length 3 , then it contains every cycle of length $3 .$
Step III. If a subgroup $N$ of $A_{n}$ contains every cycle of length 3, then it is all of $A_{n}$.
Of these, Step I is the most work. The idea is as follows. Let us say that an element $i \in\{1, \ldots, n\}$ is "moved by $\sigma$ " if $\sigma(i) \neq i$; in the contrary case we will say that $i$ is "fixed by $\sigma^{\prime \prime}$; and let us think of a permutation as "small" if it moves few elements of $\{1, \ldots, n\}$. If $x$ is any element of $N$ and $\sigma$ a "small" even permutation, then the conjugate $\sigma \times \sigma^{-1}$, which by normality of $N$ also lies in $N$, will differ only "slightly" from x; i.e., will agree with x except on a small number of elements. Hence if we "divide" the former element by the latter, the resulting permutation, $\left(\sigma x \sigma^{-1}\right) x^{-1}$, will be relatively "small". By applying this principle with a little ingenuity, we will be able to get from an arbitrarily "large" permutation down to a cycle of length $3 .$
Let us begin with a lemma making explicit the computational trick sketched above.
Lemma 1. Let $x, \sigma \in A_{n}$. Then every element $i \in\{1, \ldots, n\}$ that is moved by $\sigma x \sigma^{-1} x^{-1}$ is either an element moved by $\sigma$, or the image under $x$ of an element moved by $\sigma$.
Proof. Suppose $i \in\{1, \ldots, n\}$ is neither an element moved by $\sigma$ nor the image under $x$ of such an element. The latter condition implies that the element whose image under $x$ is $i$, namely $x^{-1}(i)$, is not moved by $\sigma$, hence is not moved by $\sigma^{-1}$ either. Hence we get $\sigma x \sigma^{-1} x^{-1}(i)=\sigma\left(x\left(\sigma^{-1}\left(x^{-1}(i)\right)\right)\right)=$ $\sigma\left(x\left(x^{-1}(i)\right)\right)=\sigma(i)=i$ as required. (The second step, i.e., the second " $=^{\prime \prime}$, uses the fact that $x^{-1}(i)$ is not moved by $\sigma$, the third step uses the identity $\mathrm{xx}^{-1}=\iota$, and the last step, the assumption that $i$ is not moved by $\sigma$.)
Now let $N$ be any nontrivial normal subgroup of $A_{n}$. We shall apply the above lemma in each of a series of cases, to show that in every case, $N$ contains a cycle of length 3 . We shall also use repeatedly the fact that if $x \in N$ and $\sigma \in A_{n}$, then $\sigma x \sigma^{-1} x^{-1}$ is the product of the two elements $\sigma x \sigma^{-1}$ and $x^{-1}$ of $N$, and hence also a member of $N$.
Case A. $N$ contains an element $x$ whose expression as a product of disjoint cycles involves at least one cycle of length greater than $3 .$
Let such a cycle be $\left(a_{1}, a_{2}, a_{3}, a_{4}, \ldots, a_{r}\right)$. (Here $r$ may equal 4 , so the " $\ldots, a_{r}$ " may be empty; but since $r>3, a_{1}, a_{2}, a_{3}, a_{4}$ are distinct.) Thus, $x=\left(a_{1}, a_{2}, a_{3}, a_{4}, \ldots, a_{r}\right) P$, where $P$ is either the identity, or a product of cycles that do not move any of $a_{1}, \ldots, a_{r}$. Let $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$. The elements of $\{1, \ldots, n\}$ that are moved by $\sigma$ are $a_{1}, a_{2}, a_{3}$, and the images under $x$ of those elements are $a_{2}, a_{3}, a_{4}$. Hence by Lemma 1, the permutation $\sigma x \sigma^{-1} x^{-1}$ cannot move any elements but $a_{1}, a_{2}$, $a_{3}, a_{4}$. Therefore we can describe $\sigma x \sigma^{-1} x^{-1}$ by determining what it does on those four elements. Go through this calculation. You will find that $\sigma \times \sigma^{-1} x^{-1}=\left(a_{1}, a_{2}, a_{4}\right)$. Hence in this case $N$ indeed contains a cycle of length 3, as claimed.
Now if a nonidentity element $x \in N$, when expressed as a product of disjoint cycles, does not involve a cycle of length greater than 3 , then it must be a product of disjoint cycles each of length 2 or length 3 . Also note that if an element $x$ is written as a product of disjoint cycles $\sigma_{1} \ldots \sigma_{r}$, then $\sigma_{1}, \ldots, \sigma_{r}$ commute with each other; hence for any integer $d$, we have $x^{d}=\sigma_{1}^{d} \ldots \sigma_{r}^{d}$. In particular, if the $x$ we are interested in here involves both cycles of length 2 and cycles of length 3 , then $x^{2}$ involves only cycles of length 3 (since the square of a cycle of length 2 is the identity, while the square of a cycle of length 3 is another cycle of length 3); and similarly, $x^{3}$ involves only cycles of length 2. Thus by either squaring or cubing such an $x$, we can see that $N$ contains an element which is either a product of one or more disjoint cycles of length 2, or of one or more disjoint cycles of length 3. We now consider those cases:
Case B. $N$ contains an element $x$ which is a product of one or more disjoint cycles of length 3. If $x$ is a single cycle of length 3 , we have what we want. In the contrary case, let us write it as $\left(a_{1}, a_{2}, a_{3}\right)\left(a_{4}, a_{5}, a_{6}\right) P$, where $P$ is again either the identity or a product of cycles that do not move any of $a_{1}, \ldots, a_{6}$. In this case, let us take $\sigma=\left(a_{2}, a_{3}, a_{4}\right)$. Using Lemma 1, we find that $\sigma \times \sigma^{-1} x^{-1}$ can only move some subset of $\left\{a_{1}, \ldots, a_{5}\right\}$. Calculating, we find that $\sigma \times \sigma^{-1}{x}^{-1}=\left(a_{1}, a_{4}, a_{2}, a_{3}, a_{5}\right), a$ cycle of length 5. A pplying the result of Case A, we conclude that $N$ also contains a cycle of length 3 .
Case C. N contains an element $x$ which is a product of one or more disjoint cycles of length 2. In this case there must be at least two such cycles, since a single cycle of length 2 is an odd permutation, while $N$ was assumed a subgroup of the group $A_{n}$ of even permutations. Hence let us write $x=$ $\left(a_{1}, a_{2}\right)\left(a_{3}, a_{4}\right) P$, with $P$ as before. Taking $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$, we find that $\sigma \times \sigma^{-1} x^{-1}=$ $\left(a_{1}, a_{3}\right)\left(a_{2}, a_{4}\right)$.
This is still a product of cycles of length 2, so have we accomplished anything? Y es, we have gotten rid of $P$, and thus have a member of $N$ which moves only 4 elements of $\{1, \ldots, n\}$. Since $n \geq 5$, this means that at least one element is fixed by this permutation. And this puts us in our final case:
Case D. $N$ contains an element $x$ such that the expression for $x$ as a product of disjoint cycles involves a cycle $\left(a_{1}, a_{2}\right)$ of length 2 , and such that $x$ fixes at least one element $a_{3}$. In this situation, let us write $x=\left(a_{1}, a_{2}\right) P$ where $P$ moves none of $a_{1}, a_{2}, a_{3}$, and let $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$. By Lemma 1, $\sigma X \sigma^{-1} x^{-1}$ can move only some subset of $\left\{a_{1}, a_{2}, a_{3}\right\}$. Computing its action on this set, you will find that it equals $\left(a_{1}, a_{3}, a_{2}\right)$, a cycle of length 3 . This completes our proof that in every case, $N$ contains such an element.
For Step II of our proof, we must show that if $N$ contains a cycle $\left(a_{1}, a_{2}, a_{3}\right)$ of length 3 , it contains every cycle $\left(b_{1}, b_{2}, b_{3}\right)$ of that length. To do this, we will need an explicit formula for the result of conjugating a cycle of length 3 by any other permutation $\sigma$. I claim that such a formula is
$$
\sigma\left(a_{1}, a_{2}, a_{3}\right) \sigma^{-1}=\left(\sigma\left(a_{1}\right), \sigma\left(a_{2}\right), \sigma\left(a_{3}\right)\right) \text {. }
$$
Indeed, it is easy to see that the two sides agree on the three elements $\sigma\left(a_{1}\right), \sigma\left(a_{2}\right)$ and $\sigma\left(a_{3}\right)$. If $i$ is not one of these elements, then $\sigma^{-1}(i)$ is not one of $a_{1}, a_{2}$ or $a_{3}$, hence it is not moved by the cycle $\left(a_{1}, a_{2}, a_{3}\right)$, so $\sigma\left(a_{1}, a_{2}, a_{3}\right) \sigma^{-1}(i)=\sigma \sigma^{-1}(i)=i$; so the left-hand side of the above display agrees with the right-hand side on such elements i as well.
Hence if our subgroup $N$ contains a cycle $\left(a_{1}, a_{2}, a_{3}\right)$ and we want to show it to contain some other cycle of length $3,\left(b_{1}, b_{2}, b_{3}\right)$, the obvious thing to do is find a permutation $\sigma$ such that $\sigma\left(a_{1}\right)=b_{1}$, $\sigma\left(a_{2}\right)=b_{2}, \sigma\left(a_{3}\right)=b_{3}$; which we can do, roughly speaking, by writing $\sigma=\left(\begin{array}{lllll}\cdots & a_{1} & \cdots & a_{2} & \cdots & a_{3} & \mathrm{~b}_{2} & \mathrm{~b}_{2} & \mathrm{~b}_{3}\end{array}\right)$ and filling in the remaining entries of the bottom row in any way such that each integer from 1 to $n$ gets used exactly once. (I say "roughly speaking" because the way I have written this permutation assumes $a_{1}<a_{2}<a_{3}$. When that is not so, the columns shown will appear in a different order.)
This will indeed give us a permutation $\sigma$ such that $\sigma\left(a_{1}, a_{2}, a_{3}\right) \sigma^{-1}=\left(b_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}\right)$; and if this $\sigma$ is even, i.e., is a member of $A_{n}$, that will show that the normal subgroup $N$ of $A_{n}$, in addition to $\left(a_{1}, a_{2}, a_{3}\right)$, also contains $\left(b_{1}, b_{2}, b_{3}\right)$. But what if $\sigma$ is odd? There are various ways one can "cure" this; the one we will use is to let $\sigma^{\prime}=\left(b_{2}, b_{3}\right) \sigma$. Since $\left(b_{2}, b_{3}\right)$ is, like $\sigma$, odd, $\sigma^{\prime}$ will be even. It will again send the three elements $a_{1}, a_{2}, a_{3}$ to $b_{1}, b_{2}, b_{3}$, but in a different order; and we see that $\sigma^{\prime}\left(a_{1}, a_{2}, a_{3}\right) \sigma^{\prime-1}=\left(b_{1}, b_{3}, b_{2}\right)$. Now the square of this cycle is the desired cycle $\left(b_{1}, b_{2}, b_{3}\right) ;$ so again, if $N$ contains $\left(a_{1}, a_{2}, a_{3}\right)$, it also contains $\left(b_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}\right)$, completing Step II.
For Step III, we must prove that the cycles of length 3 together generate $A_{n}$. A key fact will be
Lemma 2. If $x$ is a permutation which moves at least three elements, then there exists a cycle $\sigma$ of length 3 such that $\sigma^{-1} x$ moves fewer elements than $x$ does.
Proof. Let $a_{1}$ be an element moved by $x$. If we write $a_{2}=x\left(a_{1}\right)$, this is a second element moved by $x$. Since there are at least three elements moved by $x$, we can choose a third such element, $a_{3}$. Letting $\sigma=\left(a_{1}, a_{2}, a_{3}\right)$, we observe that $\sigma^{-1} x$ does not move any elements that are not moved by $x$ (since neither $x$ nor $\sigma^{-1}$ does); hence if we can prove that $\sigma^{-1} x$ fixes some element that is moved by $x$, we will have the desired conclusion. And indeed, it is immediate from the way we defined $\sigma$ that $\sigma^{-1} x$ fixes $a_{1}$.
Since the only permutations that move fewer than three elements are the identity and the transpositions, we see that every nonidentity element of $A_{n}$ moves at least three elements. Hence if $x$ is a nonidentiy element of $A_{n}$, we can use the above lemma to find a cycle $\sigma_{1}$ of length 3 such that $\sigma_{1}^{-1} x$ moves fewer elements than $x$. If $\sigma_{1}^{-1} x$ is not the identity, we can similarly use the lemma to find a cycle $\sigma_{2}$ of length 3 such that $\sigma_{2}^{-1} \sigma_{1}^{-1} x$ moves fewer elements than $\sigma_{1}^{-1} x$, and so forth. This process must eventually stop, so we must eventually get an expression for the identity permutation as
$$
\sigma_{r}^{-1} \ldots \sigma_{1}^{-1} x
$$
where $\sigma_{1}, \ldots, \sigma_{r}$ are cycles of length 3. This condition can be written $\left(\sigma_{1} \ldots \sigma_{r}\right)^{-1} x=\iota$, equivalently, $x=\sigma_{1} \ldots \sigma_{r}$, showing that the general element $x$ of $A_{n}$ is indeed a product of cycles of length $3 .$
In summary, we have shown that for $n \geq 5$, any normal subgroup $N$ of $A_{n}$ which contains a nonidentity element contains a cycle of length 3, from this that N contains all cycles of length 3, and from this that $N$ contains all elements of $A_{n}$. Thus $A_{n}$ has no proper nontrivial normal subgroup; i.e., it is simple.