Show that the set of permutations $\{\sigma \mid \sigma(i)\le i \text{ for }1\le i\le 6 \}$ is a subgroup of $S_6$.
I am stuck at how I am supposed to reason about this question. I'd assume I'd first check closure, then the group axioms, but how am I to deal with the huge variety of choices of the image? Is there supposed to be a really obvious way to show closure that I am failing to see?
Let us call a permutation $\sigma$ good if $\forall i:\sigma(i)\le i$. Consider composition $\sigma\circ\tau$, where $\sigma$ and $\tau$ are good permutations from $S_6$. Since $\tau$ is good, $\tau(i)\le i$. Since $\sigma$ is good, $\sigma(\tau(i))\le\tau(i)\le i$. Therefore, $\sigma\circ\tau$ is good. There is at least one good permutation, namely a trivial one. So good permutations form a subgroup.
The fact that there is only one good permutation could help, but we didn’t use it.
We used the finite subgroup test though.