## Tag Archives: cycle notation

### Alt(n) is generated by the set of all 3-cycles

Prove that $A_n$ is generated by the set of all 3-cycles for $n \geq 3$.

Let $n \geq 3$ and let $T = \{ (a\ b\ c) \ |\ 1 \leq a,b,c \leq n \}$ be the set of 3-cycles in $A_n$.

Note that $A_n$ contains $T$, so that $\langle T \rangle \leq A_n$.

Recall that $A_n$ consists of all permutations which can be written as an even product of transpositions; more specifically, $A_n$ is generated by the set of all products of two distinct 2-cycles. (Here we use the fact that $n \geq 3$.) Each product $\sigma$ of two 2-cycles has one of two forms.

If $\sigma = (a\ b)(c\ d)$, note that $\sigma = (a\ c\ b)(a\ c\ d)$.

If $\sigma = (a\ b)(a\ c)$, note that $\sigma = (a\ c\ b)$.

Thus $A_n \leq \langle T \rangle$, hence $A_n = \langle T \rangle$.

### Compute the size of each conjugacy class in Sym(n)

This exercise gives a formula for the size of each conjugacy class in $S_n$. Let $\sigma$ be a permutation in $S_n$ and let $m_1,m_2,\ldots,m_s$ be the distinct integers which appear in the cycle shape of $\sigma$ (including 1-cycles). Let $k_1,k_2,\ldots,k_s$ be the multiplicity of each $m_i$ in the cycle shape of $\sigma$. (That is, $\sum_{i=1}^s k_im_i = n$.) Prove that the number of conjugates of $\sigma$ is $n!/(\prod_{i=1}^s k_i! m_i^{k_i})$.

The conjugates of $\sigma$ are precisely those permutations having the same cycle type. There are $n!$ different ways we can “fill in” a permutation having the same cycle shape as $\sigma$. In doing so, we overcount due to two factors: each cycle of length $m_i$ can be cyclically permuted $m_i$ times and the cycles of length $m_i$ can be permuted among themselves (with $k_i!$ ways to do this for each $i$) without changing the permutation.

### For odd n, the n-cycles in Alt(n) comprise two conjugacy classes of equal size

Prove that if $n$ is odd then the set of all $n$-cycles consists of two conjugacy classes of equal size in $A_n$.

Let $n$ be an odd positive integer. Then any $n$-cycle is an even permutation, so that the set (hence conjugacy class) $K \subseteq S_n$ consisting of all $n$-cycles is contained in $A_n$. Note that the cycle type of any $\sigma \in K$ is $n$; that is, the cycle type of $\sigma$ consists of distinct odd integers. Thus, by this previous exercise, $K$ consists of two $A_n$-conjugacy classes of equal size.

### Find conjugators in Sym(5)

Let $\sigma = (1\ 2\ 3\ 4\ 5) \in S_5$. Find an element $\tau$ that satisfies each of the following equations: (1) $\tau\sigma\tau^{-1} = \sigma^2$, (2) $\tau\sigma\tau^{-1} = \sigma^{-1}$, and (3) $\tau\sigma\tau^{-1} = \sigma^{-2}$.

1. If $\tau = (2\ 3\ 5\ 4)$, then $\tau\sigma\tau^{-1} = (2\ 3\ 5\ 4)(1\ 2\ 3\ 4\ 5)(2\ 4\ 5\ 3)$ $= (1\ 3\ 5\ 2\ 4) = \sigma^2$.
2. If $\tau = (2\ 5)(3\ 4)$, then $\tau\sigma\tau^{-1} = (2\ 5)(3\ 4)(1\ 2\ 3\ 4\ 5)(2\ 5)(3\ 4)$ $= (1\ 5\ 4\ 3\ 2) = \sigma^{-1}$.
3. If $\tau = (2\ 4\ 5\ 3)$, then $\tau\sigma\tau^{-1} = (2\ 4\ 5\ 3)(1\ 2\ 3\ 4\ 5)(2\ 3\ 5\ 4)$ $= (1\ 4\ 2\ 5\ 3) = \sigma^{-2}$.

### Compute the orbits, cycle decompositions, and stabilizers of some group actions of Sym(3)

Repeat the previous example with each of the following actions.

1. $S_3$ acting on the set $A = \{ (i,j,k) \ |\ 1 \leq i,j,k \leq 3 \}$ by $\sigma \cdot (i,j,k) = (\sigma(i), \sigma(j), \sigma(k))$
2. $S_3$ acting on the set $B$ of all nonempty subsets of $\{1,2,3\}$ by $\sigma \cdot X = \sigma[X]$.

• For $S_3$ acting on $A$:
• We compute the orbits as follows.
• $1 \cdot (1,1,1) = (2\ 3) \cdot (1,1,1) = (1,1,1)$, $(1\ 2) \cdot (1,1,1) = (1\ 2\ 3) \cdot (1,1,1) = (2,2,2)$, and $(1\ 3) \cdot (1,1,1) = (1\ 3\ 2) \cdot (1,1,1) = (3,3,3)$, so that $\mathcal{O}_1 = \{ (1,1,1), (2,2,2), (3,3,3) \}$.
• $1 \cdot (1,1,2) = (1,1,2)$, $(1\ 2) \cdot (1,1,2) = (2,2,1)$, $(1\ 3) \cdot (1,1,2) = (3,3,2)$, $(2\ 3) \cdot (1,1,2) = (1,1,3)$, $(1\ 2\ 3) \cdot (1,1,2) = (2,2,3)$ and $(1\ 3\ 2) \cdot (1,1,2) = (3,3,1)$, so that $\mathcal{O}_2 = \{ (1,1,2), (2,2,1), (3,3,2),$ $(1,1,3), (2,2,3), (3,3,1) \}$.
• $1 \cdot (1,2,1) = (1,2,1)$, $(1\ 2) \cdot (1,2,1) = (2,1,2)$, $(2\ 3) \cdot (1,2,1) = (1,3,1)$, $(1\ 2\ 3) \cdot (1,2,1) = (2,3,2)$, and $(1\ 3\ 2) \cdot (1,2,1) = (3,1,3)$, so that $\mathcal{O}_3 = \{ (1,2,1), (2,1,2), (3,2,3),$ $(1,3,1), (2,3,2), (3,1,3) \}$.
• $1 \cdot (2,1,1) = (2,1,1)$, $(1\ 2) \cdot (2,1,1) = (1,2,2)$, $(1\ 3) \cdot (2,1,1) = (2,3,3)$, $(2\ 3) \cdot (2,1,1) = (3,1,1)$, $(1\ 2\ 3) \cdot (2,1,1) = (3,2,2)$, and $(1\ 3\ 2) \cdot (2,1,1) = (1,3,3)$, so that $\mathcal{O}_4 = \{ (2,1,1), (1,2,2),$ $(2,3,3), (3,1,1),$ $(3,2,2), (1,3,3) \}$.
• $1 \cdot (1,2,3) = (1,2,3)$, $(1\ 2) \cdot (1,2,3) = (2,1,3)$, $(1\ 3) \cdot (1,2,3) = (3,2,1)$, $(2\ 3) \cdot (1,2,3) = (1,3,2)$, $(1\ 2\ 3) \cdot (1,2,3) = (2,3,1)$, and $(1\ 3\ 2) \cdot (1,2,3) = (3,1,2)$, so that $\mathcal{O}_5 = \{ (1,2,3), (2,1,3),$ $(3,2,1), (1,3,2),$ $(2,3,1), (3,1,2) \}$.

This exhausts the elements of $A$.

• We fix the labeling
 $\alpha_{1} = (1,1,1)$ $\alpha_{10} = (2,1,1)$ $\alpha_{19} = (3,1,1)$ $\alpha_{2} = (1,1,2)$ $\alpha_{11} = (2,1,2)$ $\alpha_{20} = (3,1,2)$ $\alpha_{3} = (1,1,3)$ $\alpha_{12} = (2,1,3)$ $\alpha_{21} = (3,1,3)$ $\alpha_{4} = (1,2,1)$ $\alpha_{13} = (2,2,1)$ $\alpha_{22} = (3,2,1)$ $\alpha_{5} = (1,2,2)$ $\alpha_{14} = (2,2,2)$ $\alpha_{23} = (3,2,2)$ $\alpha_{6} = (1,2,3)$ $\alpha_{15} = (2,2,3)$ $\alpha_{24} = (3,2,3)$ $\alpha_{7} = (1,3,1)$ $\alpha_{16} = (2,3,1)$ $\alpha_{25} = (3,3,1)$ $\alpha_{8} = (1,3,2)$ $\alpha_{17} = (2,3,2)$ $\alpha_{26} = (3,3,2)$ $\alpha_{9} = (1,3,3)$ $\alpha_{18} = (2,3,3)$ $\alpha_{27} = (3,3,3)$.

Now under the permutation representation $S_3 \rightarrow S_{27}$, we have

• $1 \mapsto 1$
• $(1\ 2) \mapsto (\alpha_1\ \alpha_{14})$ $(\alpha_2\ \alpha_{13})$ $(\alpha_3\ \alpha_{15})$ $(\alpha_4\ \alpha_{11})$ $(\alpha_5\ \alpha_{10})$ $(\alpha_6\ \alpha_{12})$ $(\alpha_7\ \alpha_{17})$ $(\alpha_8\ \alpha_{16})$ $(\alpha_9\ \alpha_{18})$ $(\alpha_{19}\ \alpha_{23})$ $(\alpha_{20}\ \alpha_{22})$ $(\alpha_{21}\ \alpha_{24})$ $(\alpha_{25}\ \alpha_{26})$
• $(1\ 3) \mapsto (\alpha_1\ \alpha_{27})$ $(\alpha_2\ \alpha_{26})$ $(\alpha_3\ \alpha_{25})$ $(\alpha_4\ \alpha_{24})$ $(\alpha_5\ \alpha_{23})$ $(\alpha_6\ \alpha_{22})$ $(\alpha_7\ \alpha_{21})$ $(\alpha_8\ \alpha_{20})$ $(\alpha_9\ \alpha_{19})$ $(\alpha_{10}\ \alpha_{18})$ $(\alpha_{11}\ \alpha_{17})$ $(\alpha_{12}\ \alpha_{16})$ $(\alpha_{13}\ \alpha_{15})$
• $(2\ 3) \mapsto (\alpha_2\ \alpha_3)$ $(\alpha_4\ \alpha_7)$ $(\alpha_5\ \alpha_{9})$ $(\alpha_6\ \alpha_{8})$ $(\alpha_{10}\ \alpha_{19})$ $(\alpha_{11}\ \alpha_{21})$ $(\alpha_{12}\ \alpha_{20})$ $(\alpha_{13}\ \alpha_{25})$ $(\alpha_{14}\ \alpha_{27})$ $\alpha_{15}\ \alpha_{26})$ $(\alpha_{16}\ \alpha_{22})$ $(\alpha_{17}\ \alpha_{24})$ $(\alpha_{18}\ \alpha_{23})$
• $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_{14}\ \alpha_{27})$ $(\alpha_2\ \alpha_{15}\ \alpha_{25})$ $(\alpha_3\ \alpha_{13}\ \alpha_{26})$ $(\alpha_{4}\ \alpha_{17}\ \alpha_{21})$ $(\alpha_5\ \alpha_{18}\ \alpha_{19})$ $(\alpha_6\ \alpha_{16}\ \alpha_{20})$ $(\alpha_7\ \alpha_{11}\ \alpha_{24})$ $(\alpha_8\ \alpha_{12}\ \alpha_{22})$ $(\alpha_9\ \alpha_{10}\ \alpha_{23})$
• $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_{27}\ \alpha_{14})$ $(\alpha_2\ \alpha_{25}\ \alpha_{15})$ $(\alpha_3\ \alpha_{26}\ \alpha_{13})$ $(\alpha_{4}\ \alpha_{21}\ \alpha_{17})$ $(\alpha_5\ \alpha_{19}\ \alpha_{18})$ $(\alpha_6\ \alpha_{20}\ \alpha_{16})$ $(\alpha_7\ \alpha_{24}\ \alpha_{11})$ $(\alpha_8\ \alpha_{22}\ \alpha_{12})$ $(\alpha_9\ \alpha_{23}\ \alpha_{10})$
• If $x \in \mathcal{O}_1$, then $[S_3 : \mathsf{stab}(x)] = 3$ so that $|\mathsf{stab}(x)| = 2$. In particular, consider $(3,3,3) \in \mathcal{O}_1$. Since $(1\ 2) \cdot (3,3,3) = (3,3,3)$, we have $\mathsf{stab}((3,3,3)) = \langle (1\ 2) \rangle$.

Each remaining orbit has order 6. Thus, if $x \in \mathcal{O}_k$, $2 \leq k \leq 5$, then $|\mathsf{stab}(x)| = 1$; hence $\mathsf{stab}(x) = 1$.

• For $S_3$ acting on $B$:
• We compute the orbits as follows.
• We have $1 \cdot \{1\} = (2\ 3) \cdot \{1\} = \{1\}$, $(1\ 2) \cdot \{1\} = (1\ 2\ 3) \cdot \{1\} = \{2\}$, and $(1\ 3) \cdot \{1\} = (1\ 3\ 2) \cdot \{1\} = \{3\}$. Thus $\mathcal{O}_1 = \{ \{1\}, \{2\}, \{3\} \}$.
• We have $1 \cdot \{1,2\} = (1\ 2) \cdot \{1,2\} = \{1,2\}$, $(1\ 3) \cdot \{1,2\} = (1\ 2\ 3) \cdot \{1,2\} = \{2,3\}$, and $(2\ 3) \cdot \{1,2\} = (1\ 3\ 2) \cdot \{1,2\} = \{1,3\}$. Thus $\mathcal{O}_2 = \{ \{1,2\}, \{1,3\}, \{2,3\} \}$.
• There is only one remaining element of $B$. Thus $\mathcal{O}_3 = \{ \{1,2,3\} \}$.
• We fix the labeling
 $\alpha_{1} = \{1\}$ $\alpha_{5} = \{1,3\}$ $\alpha_{2} = \{2\}$ $\alpha_{6} = \{2,3\}$ $\alpha_{3} = \{3\}$ $\alpha_{7} = \{1,2,3\}$ $\alpha_{4} = \{1,2\}$

Under the permutation representation $S_3 \rightarrow S_7$,

• $1 \mapsto 1$
• $(1\ 2) \mapsto (\alpha_1\ \alpha_2)(\alpha_5\ \alpha_6)$
• $(1\ 3) \mapsto (\alpha_1\ \alpha_3)(\alpha_4\ \alpha_6)$
• $(2\ 3) \mapsto (\alpha_2\ \alpha_3)(\alpha_4\ \alpha_5)$
• $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_2\ \alpha_3)(\alpha_4\ \alpha_6\ \alpha_5)$
• $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_3\ \alpha_2)(\alpha_4\ \alpha_5\ \alpha_6)$
• If $x \in \mathcal{O}_1$, then $[S_3 : \mathsf{stab}(x)] = 3$, so that $|\mathsf{stab}(x)| = 2$. In particular, since $(2\ 3) \cdot \{1\} = \{1\}$, we have $\mathsf{stab}(\{1\}) = \langle (2\ 3) \rangle$.

If $x \in \mathcal{O}_2$, we similarly have $|\mathsf{stab}(x)| = 2$. In particular, since $(2\ 3) \cdot \{2,3\} = \{2,3\}$, we have $\mathsf{stab}(\{2,3\}) = \langle (2\ 3) \rangle$.

Now $[S_3 : \mathsf{stab}(\{1,2,3\})] = 1$, so that $\mathsf{stab}(\{1,2,3\}) = S_3$.

### Compute the orbits, cycle decompositions, and stabilizers of some given group actions of Sym(3)

Let $S_3$ act on the set $A = \{ (i,j) \ |\ 1 \leq i,j \leq 3 \}$ by $\sigma \cdot (i,j) = (\sigma(i), \sigma(j))$.

1. Find the orbits of $S_3$ on $A$.
2. Fix a labeling of the elements of $A$. For each $\sigma \in S_3$, find the cycle decomposition of $\sigma$ under the permutation representation $S_3 \rightarrow S_9$.
3. For each orbit $\mathcal{O}$ of this action, choose some $a \in \mathcal{O}$ and compute $\mathsf{stab}(a)$.

1. We have $1 \cdot (1,1) = (2\ 3) \cdot (1,1) = (1,1)$, $(1\ 2) \cdot (1,1) = (1\ 2\ 3) \cdot (1,1) = (2,2)$, and $(1\ 3) \cdot (1,1) = (1\ 3\ 2) \cdot (1,1) = (3,3)$. Thus the orbit containing $(1,1)$ is $\mathcal{O}_1 = \{ (1,1), (2,2), (3,3) \}$.

We also have $1 \cdot (1,2) = (1,2)$, $(1\ 2) \cdot (1,2) = (2,1)$, $(1\ 3) \cdot (1,2) = 3,2)$, $(2\ 3) \cdot (1,2) = (1,3)$, $(1\ 2\ 3) \cdot (1,2) = (2,3)$, and $(1\ 3\ 2) \cdot (1,2) = (3,2)$. Thus the orbit containing $(1,2)$ is $\mathcal{O}_2 = \{ (1,2), (2,1), (3,2),$ $(1,3), (2,3), (3,2) \}$. These two orbits exhaust the elements of $A$.

2. Fix the labeling
 $\alpha_{1} = (1,1)$ $\alpha_{4} = (2,1)$ $\alpha_{7} = (3,1)$ $\alpha_{2} = (1,2)$ $\alpha_{5} = (2,2)$ $\alpha_{8} = (3,2)$ $\alpha_{3} = (1,3)$ $\alpha_{6} = (2,3)$ $\alpha_{9} = (3,3)$

Under the permutation representation $S_3 \rightarrow S_9$, we have

1. $1 \mapsto 1$
2. $(1\ 2) \mapsto (\alpha_1\ \alpha_5)(\alpha_2\ \alpha_4)(\alpha_3\ \alpha_6)(\alpha_7\ \alpha_8)$
3. $(1\ 3) \mapsto (\alpha_1\ \alpha_9)(\alpha_2\ \alpha_8)(\alpha_3\ \alpha_7)(\alpha_4\ \alpha_6)$
4. $(2\ 3) \mapsto (\alpha_2\ \alpha_3)(\alpha_4\ \alpha_7)(\alpha_5\ \alpha_9)(\alpha_6\ \alpha_8)$
5. $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_5\ \alpha_9)(\alpha_2\ \alpha_6\ \alpha_7)(\alpha_3\ \alpha_4\ \alpha_8)$
6. $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_9\ \alpha_5)(\alpha_2\ \alpha_7\ \alpha_6)(\alpha_3\ \alpha_8\ \alpha_4)$
3. If $\sigma \in \mathcal{O}_1$, we have $[S_3 : \mathsf{stab}(\sigma)] = 3$, so that $|\mathsf{stab}(\sigma)| = 2$. Consider $(2,2) \in \mathcal{O}_1$; since $(1\ 3) \cdot (2,2) = (2,2)$, we have $\mathsf{stab}((2,2)) = \langle (1\ 3) \rangle$.

If $\sigma \in \mathcal{O}_2$, then $[S_3 : \mathsf{stab}(\sigma)] = 6$. Hence $|\mathsf{stab}(\sigma)| = 1$, and we have $\mathsf{stab}(\sigma) = 1$.

### Find a generating set for Sym(p)

Let $p$ be a prime. Show that $S_p = \langle \sigma, \tau \rangle$ where $\sigma$ is any transposition and $\tau$ any $p$-cycle.

Let $\sigma = (a_1\ a_2)$ and $\tau = (a_1\ b_2\ \ldots\ b_p)$. (We have $a_2 = b_i$ for some $i$.) By a previous exercise, $\tau^k(a_1) = a_2$ for some $k$. Because $\tau$ has prime order, $\langle \sigma, \tau \rangle = \langle \sigma, \tau^k \rangle$. Relabeling the elements $\{ 1, \ldots, n \}$, by the previous exercise we have $S_p = \langle \sigma, \tau^k \rangle = \langle \sigma, \tau \rangle$.

### Determine whether a given permutation is even or odd

In these two previous exercises, we were asked to find the cycle decompositions of some permutations. Write each of those permutations as a product of transpositions and determine whether each is odd or even.

• $(1\ 3\ 5)(2\ 4)$ $= (1\ 5)(1\ 3)(2\ 4)$ is odd.
• $(1\ 5)(2\ 3)$ is even.
• $(1\ 5\ 3)$ $= (1\ 3)(1\ 5)$ is even.
• $(2\ 5\ 3\ 4)$ $= (2\ 4)(2\ 3)(2\ 5)$ is odd.
• $(1\ 2\ 4\ 3)$ $= (1\ 3)(1\ 4)(1\ 2)$ is odd.
• $(1\ 13\ 5\ 10)(3\ 15\ 18)(4\ 14\ 11\ 7\ 12\ 9)$ $= (1\ 10)(1\ 5)(1\ 13)(3\ 18)(3\ 15)(4\ 9)(4\ 12)(4\ 7)(4\ 11)(4\ 14)$ is even.
• $(1\ 14)(2\ 9\ 15\ 13\ 4)(3\ 10)(5\ 12\ 7)(8\ 11)$ $= (1\ 14)(2\ 4)(2\ 13)(2\ 15)(2\ 9)(3\ 10)(5\ 7)(5\ 12)(8\ 11)$ is odd.
• $(1\ 5)(3\ 8\ 15)(4\ 11\ 12)(7\ 9\ 14)(10\ 13)$ $= (1\ 5)(3\ 15)(3\ 8)(4\ 12)(4\ 11)(7\ 14)(7\ 9)(10\ 13)$ is even.
• $(1\ 11\ 3)(2\ 4)(5\ 9\ 8\ 7\ 10\ 15)(13\ 14)$ $= (1\ 3)(1\ 11)(2\ 4)(5\ 15)(5\ 10)(5\ 7)(5\ 8)(5\ 9)(13\ 14)$ is odd.
• $(1\ 4)(2\ 9)(3\ 13\ 12\ 15\ 11\ 5)(8\ 10\ 14)$ $= (1\ 4)(2\ 9)(3\ 5)(3\ 11)(3\ 15)(3\ 12)(3\ 13)(8\ 14)(8\ 10)$ is odd.
• $(1\ 2\ 15\ 8\ 3\ 4\ 14\ 11\ 12\ 13\ 7\ 5\ 10)$ $= (1\ 10)(1\ 5)(1\ 7)(1\ 13)(1\ 12)(1\ 11)(1\ 14)(1\ 4)(1\ 3)$ $\circ (1\ 8)(1\ 15)(1\ 2)$ is even.

### Exhibit a generating set of Sym(4) consisting of two 4-cycles

Prove that $S_4 = \langle (1\ 2\ 3\ 4), (1\ 2\ 4\ 3) \rangle$.

We have

1. $(1\ 2\ 3\ 4)^2 = (1\ 3)(2\ 4)$
2. $(1\ 2\ 3\ 4)^3 = (1\ 4\ 3\ 2)$
3. $(1\ 2\ 3\ 4)^4 = 1$
4. $(1\ 2\ 4\ 3)^2 = (1\ 4)(2\ 3)$
5. $(1\ 2\ 4\ 3)^3 = (1\ 3\ 4\ 2)$
6. $(1\ 2\ 3\ 4)(1\ 2\ 4\ 3) = (1\ 3\ 2)$
7. $(1\ 2\ 4\ 3)(1\ 2\ 3\ 4) = (1\ 4\ 2)$
8. $(1\ 3\ 2)^2 = (1\ 2\ 3)$
9. $(1\ 4\ 2)^2 = (1\ 2\ 4)$
10. $(1\ 3\ 2)(1\ 2\ 4) = (2\ 4\ 3)$
11. $(1\ 4\ 2)(1\ 2\ 3) = (2\ 3\ 4)$

So that $|\langle (1\ 2\ 3\ 4), (1\ 2\ 4\ 3) \rangle| \geq 13$. By Lagrange’s Theorem, the order of a subgroup of a finite group must divide the order of the group; thus $|\langle (1\ 2\ 3\ 4), (1\ 2\ 4\ 3) \rangle| = 24$ and we have $\langle (1\ 2\ 3\ 4), (1\ 2\ 4\ 3) \rangle = S_4$.

### Sym(3) is generated by any pair of distinct 2-cycles

Prove that the subgroup of $S_3$ generated by any two distinct elements of order 2 is all of $S_3$.

The only elements of order 2 in $S_3$ are 2-cycles. Let $(a\ b)$, $(c\ d)$ be arbitrary distinct elements of order 2 in $S_3$; these cycles cannot be disjoint, so we assume without loss of generality that $a = d$, so that in fact $(a\ b)$ and $(a\ c)$ can represent arbitrary distinct elements of order 2 in $S_3$. We can compute $\langle (a\ b), (a\ c) \rangle$ explicitly; $(a\ b)^2 = 1$, $(a\ b)(a\ c) = (a\ c\ b)$, $(a\ c)(a\ b) = (a\ b\ c)$, and $(a\ b)(a\ c)(a\ b) = (b\ c)$. Thus we have generated all of $S_3$.