I'm trying to go through the process of determining whether a set of permutations is a subgroup. I was hoping someone could look over the logic of this answer for completeness and soundness of logic. I think I understand, ultimately, how to check this, but I worry that I'm 'reinventing the wheel' when there's a much easier, more intuitive approach. (In other words, I would never want to replicate something like this on a timed exam.)
We have the set: $\{\text{id}, (12345), (13524), (14253), (15432)\}$.
We need to check four properties:
(a) Associativity: This is trivially the case, as it holds for all permutations and for the group, $S_5$, so it will surely hold for a subset of that group, regardless of whether those permutations form a subgroup.
(b) Identity: The identity permutation is indeed included in this set.
(c) Closure: I think the only way to check this is to take all possible compositions of the permutations in the group and ensure that they produce elements in the group.
The identity, clearly, will produce any of the permutations we compose it with, and it commutes, so we can for the moment omit that from our verification process. Let's denote these permutations as:
\begin{align*}
& (12345) = \sigma \\
& (13524) = \beta \\
& (14253) = \tau \\
& (15432) = \gamma
\end{align*}
I then can check each of these. Upon working these out on scrap paper (the easiest way would perhaps be with a group table), I get:
\begin{align*}
& \tau \sigma = (15432) = \gamma \\
& \sigma \tau = (15432) = \gamma \\
& \gamma \sigma = \text{id} \\
& \sigma \gamma = \text{id} \\
& \beta \sigma = (14253) = \tau \\
& \sigma \beta = (14253) = \tau \\
& \tau \sigma = (15432) = \gamma \\
& \sigma \sigma = (13524) = \beta \\
& \beta \tau = \text{id} \\
& \tau \beta = \text{id} \\
& \gamma \tau = (15432) = \beta \\
& \tau \gamma = (13524) = \beta \\
& \tau \tau = (12345) = \sigma \\
& \beta \gamma = (12345) = \sigma \\
& \gamma \beta = (12345) = \sigma \\
& \beta \beta = (15432) = \gamma \\
& \gamma \gamma = (14253) = \tau
\end{align*}
So, we can see from this that the group is closed under composition of permutations.
Inverses: The inverse of the identity is clearly the identity. The inverse of $\beta$ is $\tau$, and thus the identity of $\tau$ is $\beta$. The inverse of $\gamma$ is $\sigma$, and thus the inverse of $\sigma$ is $\gamma$. Thus, every element of this group has an inverse.
So, it seems–unless I made a mistake–that these four conditions check out. My questions, then, are:
(a) How does this look?
(b) Is there an easier, more intuitive way to approach such a problem other than writing out the group table?
Best Answer
You have the right idea for needing to check associativity, closure, and inverses. Assuming there's no mistake in your computations, your reasoning is solid, but know that there is a trick to drastically cut down on the amount of work required:
Let $S$ be the set in question. If $S$ is indeed a subgroup, we'll have $S \cong \mathbb{Z}_5$—any group of prime order is isomorphic to the cyclic group of that order, and any non-identity element of such a group serves as a generator$^\dagger$. So simply choose any non-identity permutation $\pi \in S$ and repeatedly compose it with itself to see if you get all the other elements in $S$. If you do, then we'll have $S = \{\text{id}, \pi, \pi^2, \pi^3, \pi^4 \}$. If you don't, then $S$ won't be a subgroup.
If you can realize all the elements of $S$ as compositions of one of its elements as such, then verifying the subgroup requirements becomes much easier. For example, we'll have closure since $\pi^n \circ \pi^m = \pi^{n+m \pmod{5}}$. Using this trick, you'll be able to answer the question with no more than four computations, as opposed to $17$ of them. Even more good news: it looks like you already have the rest of the work required for finishing up here.
$^\dagger$ You can prove this with Lagrange's theorem.