4
$\begingroup$

I'm studying for a midterm for my online algebra course and am taking a practice exam. Unfortunately, the exam has no accompanying solutions so I was hoping to appeal to the kind folks here for validation/clarification about, specifically, a multi-part question regarding permutations.


Part a

Let $\sigma = (123)(45)(6789)$ and $\tau = (13)(578)(49) \in S_9$. Compute the order of $\sigma$.

Here the order of a permutation that's comprised of disjoint cycles is the least common multiple of the orders of the cycles. Thus we have $$LCM(3,2,4) = \boxed{12}$$


Part b

Is $\sigma$ even or odd

Since $\sigma$ is a $12$-cycle and $12$ is even that means $\sigma$ is $\boxed{\text{odd}}$.


Part c

Compute the composition $\tau \sigma$

$\tau \sigma = \boxed{(12)(475968)}$


Part d

Is the $2$-cycle $(12)$ in the subgroup $H = \langle \sigma, \tau \rangle < S_9$ generated by $\sigma$ and $\tau$?

For this I am a bit stuck. First I am slightly confused by the notation $\langle \sigma, \tau \rangle$. I am used to seeing notation like $\langle g \rangle$ which simply means the cyclic subgroup generated by the element $g$. So, in this case I'm assuming $\langle \sigma, \tau \rangle$ means "the cyclic subgroup generated by the elements $\sigma$ and $\tau$"? What exactly does that look like. For instance, I know $\langle g \rangle = \{g^n \mid n \in \mathbb{Z} \} = G$ is the definition of the cyclic subgroup generated by the element $g$ but for $\langle \sigma, \tau \rangle$ would it be something like $\{(\sigma \tau)^n \mid n \in \mathbb{Z} \}$ or maybe $\{\sigma^n \tau^m \mid n,m \in \mathbb{Z} \}$?

If this is true then wouldn't the question be trivial since if $(12) \in H$ then since $H$ is generated by $\sigma$ and $\tau$ then every element in the subgroup is generated by these $2$ elements as well? Or could it be the case that $(12)$ is only generated by only one of the $2$ elements? I'm unsure how exactly to proceed on this question.

$\endgroup$
1
  • $\begingroup$ Please ask one question at a time. $\endgroup$
    – Shaun
    Jul 6 at 10:54

4 Answers 4

3
$\begingroup$

For b:

$\sigma$ has order $12$, but it is not a $12$-cycle. $$ (abc\cdots z) = (ab)(ac)\cdots(az) $$ so $$ \sigma =(12)(13)(45)(67)(68)(69) $$ is the product of $6$ transpositions, so is even.

For c:

Your calculation is probably correct if your text multiplies permuations in cycle notation from right to left. (I haven't checked.)

If (as is more common) it's left to right as written then $\tau$ sends $1$ to $3$ and $\sigma$ sends $3$ to $1$ so $\tau \sigma$ starts $(1) \cdots$ and not $(12)\cdots$.

For d:

$\langle \sigma , \tau \rangle$ is not cyclic. It consists of all the elements you can get by multiplying copies of $\sigma$ and $\tau$ and their inverses in any order.

Hint: what is the parity of $\tau$? of $(12)$?

$\endgroup$
1
$\begingroup$

Part (a) is good.

For part (b), $\sigma$ is not a $12$-cycle. It's the product of disjoint cycles of lengths $3$, $2$, and $4$, meaning a minimum length expression for $\sigma$ in terms of transpositions involves $$ (3-1) + (2-1) + (4-1) = 2 + 1 + 3 = 6 $$ transpositions, which is an even number, hence $\sigma$ is an even permutation. By the way, you can easily write down such an expression: $$ \color{red}{(1\,2\,3)} \color{green}{(4\,5)} \color{blue}{(6\,7\,8\,9)} = \color{red}{(1\,2) (2\,3)} \color{green}{(4\,5)} \color{blue}{(6\,7) (7\,8) (8\,9).} $$ (Surely, you're familiar with the theorem that any expression for $\sigma$ in terms of transpositions will have the same parity, which is why the sign is well-defined.)

Part (c) looks good, as well.

For part (d), here's the correct definition for $H = \langle \sigma, \tau \rangle$, the subgroup generated by the two elements $\sigma$ and $\tau$: the smallest subgroup of $G = S_9$ containing those elements. This means that you can write down any word in those symbols and their inverses, e.g. $\sigma^2 \tau^{-1} \sigma^{-4} \tau^3 \tau \sigma$, because by definition of a subgroup, they all have to be in $H$.

There's an easy way to see that the transposition $(1\,2)$ is not in $H$ though. Both $\sigma$ and $\tau$ are even transpositions. Hence, any element of $H$, will be even as well, but the single transposition is obviously odd. As a consequence $H \leq A_9$, the alternating subgroup of $S_9$, consisting of all even permutations, and $(1\,2) \notin A_9$.

Why? When you multiply permutations in $G = S_n$ (for any $n \geq 0$), length adds modulo $2$: $$ \ell(g_1 g_2) \equiv \ell(g_1) + \ell(g_2) \bmod{2} $$ for any $g_1, g_2 \in G$, where $\ell: G \to \mathbb{Z}_{\geq0}$ is the length function. Hence, $\operatorname{sgn}: S_n \to C_2$ is a homomorphism from the symmetric group to the (multiplicative) cyclic group of order $2$: $$ \operatorname{sgn}(g_1 g_2) = (-1)^{\ell(g_1 g_2)} = (-1)^{\ell(g_1)} \, (-1)^{\ell(g_2)} = \operatorname{sgn}(g_1) \, \operatorname{sgn}(g_2). $$

$\endgroup$
1
$\begingroup$

$\sigma$ and $\tau$ are even, whereas $(12)$ is odd.

It appears that you need to be introduced to the notion of a word. A word in $\sigma $ and $\tau$ will be any product of powers of $\sigma$ or $\tau$ and powers of their inverses.

Now, it's fairly easy that any word in even permutations is even. (In mathematical parlance the sgn function ${\bf\text{sgn}}:G\to\{\pm1\}$ is a homomorphism.)

Thus, $(12)\not \in\langle \sigma, \tau\rangle $.

$\endgroup$
1
$\begingroup$

The easy way to see that your reasoning can't be correct for part b is to note that if you remove the $(4~5)$ transposition from $\sigma$, the new group element $(1~2~3)(6~7~8~9)$ still would have order $12$ but (because you've removed a single transposition) would necessarily have the opposite parity. In other words, two different elements of order $12$ have different parities, so merely determining the element's order can't tell you its parity.

The correct analysis is to note that $\sigma$ is a product of a $3$-cycle, a $2$-cycle, and a $4$-cycle. The $3$-cycle is even, and the other two cycles are odd, so $\sigma$ is the product of an even element and two odd elements, and therefore is even.

The notation $\langle \sigma, \tau \rangle$ means the smallest subgroup (not necessarily cyclic) of $S_9$ that contains both $\sigma$ and $\tau$. You can simplify the problem by noting that the set $\{1, 2, 3 \}$ is invariant when permuted by both $\sigma$ and $\tau$, so you can focus on products of $\sigma$ and $\tau$ that result in the transposition $(1~3)$ when acting on that set. All you need to figure out then is whether those products can possibly leave the other elements $\{4, 5, 6, 7, 8, 9 \}$ fixed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.