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.
Best Answer
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 $(12)$?