Hw11 Problem 12

### Problem 12

(*) Prove that $A_{n}$ is simple for $n \geq 5$ by following the steps outlined in problem 39 of section 15 in the book.

Solution

(a) Because $(a, b, c) = (a, c)(a, b)$, we see that every 3-cycle is an even permutation, and hence is in $A_n$.

(b) Let $ \sigma \in A_n$ and write $\sigma$ as a product of transpositions. The number of transpositions in the product will be even by definition of $A_n$. The product of the first two transpositions will be either of the form $(a, b)(c, d)$ or of the form $(a, b)(a, c)$ or of the form $(a, b)(a, b)$, depending on repetition of letters in the transpositions. If the form is $(a, b)(a, b)$, it can be deleted from the product altogether. As the hint shows, either of the other two forms can be expressed as a 3-cycle. We then proceed with the next pair of transpositions in the product, and continue until we have expressed $\sigma$ as a product of 3-cycles. Thus the 3-cycles generate $A_n$.

(c) Following the hint, we find that

(1)
$$(r, s, i)^2 = (r, i, s),$$
(2)
$$(r, s, j)(r, s, i)^2 = (r, s, j)(r, i, s) = (r, i, j),$$
(3)
$$(r, s, j)^{2}(r, s, i) = (r, j, s)(r, s, i) = (s, i, j),$$
(4)
$$(r, s, i)^{2}(r, s, k)(r, s, j)^{2}(r, s, i) = (r, i, s)(r, s, k)(s, i, j) = (i, j, k).$$

Now every 3-cycle either contains neither $r$ nor $s$ and is of the form $(i, j, k)$, or just one of $r$ or $s$ and is of the form $(r, i, j)$ or $(s, i, j)$, or both $r$ and $s$ and is of the form $(r, s, i)$ or $(r, i, s) = (s, r, i)$. Because all of these forms can be obtained from our special 3-cycles, we see that the special 3-cycles generate $A_n$.

(d) Following the hint and using part (c) we find that

(5)
$$((r, s)(i, j))(r, s, i)^{2}((r, s)(i, j))^{-1} = (r, s)(i, j)(r, i, s)(i, j)(r, s) = (r, s, j).$$

Thus if $N$ is a normal subgroup of $A_n$ and contains a 3-cycle, which we can consider to be $(r, s, i)$ because $r$ and $s$ could be any two numbers from $1$ to $n$ in part (c), we see that $N$ must contain all the special 3-cycles and hence be all of $A_n$ by part (c).

(e) Before making the computations in the hints of the five cases, we observe that one of the cases must hold. If Case 1 is not true and Case 2 is not true, then when elements of $N$ are written as a product of disjoint cycles, no cycle of length greater than 3 occurs, and no element of $N$ is a single 3-cycle. The remaining cases cover the possibilities that at least one of the products of disjoint cycles involves two cycles of length 3, involves one cycle of length 3, or involves no cycle of length 3. Thus all possibilities are covered, and we now turn to the computations in the hints.

$Case\ 1$. By part (d), if $N$ contains a 3-cycle, then $N = A_n$ and we are done.

$Case\ 2$. Note that $a_1, a_2, \ldots, a_r$ do not appear in $\mu$ because the product contained disjoint cycles. We have

(6)
\begin{align} \sigma^{-1}[(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}] \end{align}
(7)
\begin{align} = (a_r, \ldots, a_2, a_1)\mu^{-1}(a_1, a_2, a_3)\mu(a_1, a_2, \ldots, a_r)(a_1, a_3, a_2) \end{align}
(8)
$$= (a_1, a_3, a_r),$$

and this element is in $N$ because it is the product of $\sigma^{-1}$ and a conjugate of $\sigma$ by an element of $A_n$. Thus in this case, $N$ contains a 3-cycle and is equal to $A_n$ by part (d).

$Case\ 3$. Note that $a_1, a_2, \ldots, a_6$ do not appear in $\mu$. As in Case 2, we see that

(9)
\begin{align} \sigma^{-1}[(a_1, a_2, a_4)\sigma(a_1, a_2, a_4)^{-1}] \end{align}
(10)
\begin{align} = (a_1, a_3, a_2)(a_4, a_6, a_5)\mu^{-1}(a_1, a_2, a_4)\mu(a_4, a_5, a_6)(a_1, a_2, a_3)(a_1, a_4, a_2) \end{align}
(11)
$$= (a_1, a_4, a_2, a_6, a_3)$$

is in $N$. Thus $N$ contains a cycle of length greater than 3, and $N = A_n$ by Case 2.

$Case\ 4$. Note that $a_1, a_2$, and $a_3$ do not appear in $\mu$. Of course $\sigma^{2} \in N$ because $\sigma \in N$, so $\sigma^{2} = \mu(a_1, a_2, a_3)\mu(a_1, a_2, a_3) = (a1, a3, a2) \in N$, so $N$ contains a 3-cycle and hence $N = A_n$ as shown by part (d).

$Case\ 5$. Note that $a_1, a_2, a_3$, and $a_4$ do not appear in $\mu$. As in Case 2, we see that

(12)
\begin{align} \sigma^{-1}[(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}] \end{align}
(13)
\begin{align} = (a_1, a_2)(a_3, a_4)\mu^{-1}(a_1, a_2, a_3)\mu(a_3, a_4)(a_1, a_2)(a_1, a_3, a_2) \end{align}
(14)
$$= (a_1, a_3)(a_2, a_4)$$

is in $N$. Continuing with the hint given, we let $\alpha = (a_1, a_3)(a_2, a_4)$ and $\beta = (a_1, a_3, i)$ where $i$ is different from $a_1, a_2, a_3$, and $a_4$. Then $\beta \in A_{n}$ and $\alpha \in N$ and $N$ a normal subgroup of $A_n$ imply that $(\beta^{-1}\alpha\beta)\alpha \in N$. Computing, we find that

(15)
\begin{align} (\beta^{-1}\alpha\beta)\alpha = (a_1, i, a_3)(a_1, a_3)(a_2, a_4)(a_1, a_3, i)(a_1, a_3)(a_2, a_4) = (a_1, a_3, i). \end{align}

Thus $N = A_n$ in this case also, by part (d).

Therefore $A_{n}$ is simple for $n \geq 5$. Q.E.D.