It is good for understanding that a small number of generators does not mean a small group.

At your very earliest encounter with permutation groups, you might think that with two generators $\alpha$ and $\beta$, well, how much can you get? Especially if one or both of them are of small order? I mean, with $\alpha=(1,2)$ you just have $\alpha^2=\text{id}$, and $\beta=(1,2,\ldots,n)$ also gives you just $n$ different permutations $\beta^1,\ldots,\beta^n=\text{id}$ so "obviously" there cannot be much more, can there?

An obvious analogy is the dihedral group $D_n$, where from your two generators — a rotation, which gives you $n$ permutations, and a reflection that gives you two — you get $2n$ permutations. So you might expect not much more here — and you would be surprised.

(And a small afterthought, which is starting to wander off-topic: It may be instructive to note that the huge diffence we see here between $D_n$ and $S_n$ does not arise from the number of generators, or from their orders, nor from commutativity vs. noncommutativity. Surely, for an abelian group one is not suprised that a small number of small-order generators gives you only a small number of group elements, because order of composing the generators does not matter. But $D_n$ and $S_n$ are both nonabelian beyond the very smallest cases, yet they behave very differently.)

It is good for understanding that a small number of generators does not mean a small group.

At your very earliest encounter with permutation groups, you might think that with two generators $\alpha$ and $\beta$, well, how much can you get? Especially if one or both of them are of small order? I mean, with $\alpha=(1,2)$ you just have $\alpha^2=\text{id}$, and $\beta=(1,2,\ldots,n)$ also gives you just $n$ different permutations $\beta^1,\ldots,\beta^n=\text{id}$ so "obviously" there cannot be much more, can there?

An obvious analogy is the dihedral group $D_n$, where from your two generators — a rotation, which gives you $n$ permutations, and a reflection that gives you two — you get $2n$ permutations. So you might expect not much more here — and you would be surprised.

(And a small afterthought, which is starting to wander off-topic: It may be instructive to note that the huge difference we see here between $D_n$ and $S_n$ does not arise from the number of generators, or from their orders, nor from commutativity vs. noncommutativity. Surely, for an abelian group one is not surprised that a small number of small-order generators gives you only a small number of group elements, because order of composing the generators does not matter. But $D_n$ and $S_n$ are both nonabelian beyond the very smallest cases, yet they behave very differently. — In fact, I'm not sure how best to characterize this difference, or whether there is a meaningful general phenomenon there, or whether it is just "that's the way these two groups are".)

It is good for understanding that a small number of generators does not mean a small group.

At your very earliest encounter with permutation groups, you might think that with two generators $\alpha$ and $\beta$, well, how much can you get? Especially if one or both of them are of small order? I mean, with $\alpha=(1,2)$ you just have $\alpha^2=\text{id}$, and $\beta=(1,2,\ldots,n)$ also gives you just $n$ different permutations $\beta^1,\ldots,\beta^n=\text{id}$ so "obviously" there cannot be much more, can there?

An obvious analogy is the dihedral group $D_n$, where from your two generators — a rotation, which gives you $n$ permutations, and a reflection that gives you two — you get $2n$ permutations. So you might expect not much more here — and you would be surprised.

It is good for understanding that a small number of generators does not mean a small group.

At your very earliest encounter with permutation groups, you might think that with two generators $\alpha$ and $\beta$, well, how much can you get? Especially if one or both of them are of small order? I mean, with $\alpha=(1,2)$ you just have $\alpha^2=\text{id}$, and $\beta=(1,2,\ldots,n)$ also gives you just $n$ different permutations $\beta^1,\ldots,\beta^n=\text{id}$ so "obviously" there cannot be much more, can there?

An obvious analogy is the dihedral group $D_n$, where from your two generators — a rotation, which gives you $n$ permutations, and a reflection that gives you two — you get $2n$ permutations. So you might expect not much more here — and you would be surprised.

(And a small afterthought, which is starting to wander off-topic: It may be instructive to note that the huge diffence we see here between $D_n$ and $S_n$ does not arise from the number of generators, or from their orders, nor from commutativity vs. noncommutativity. Surely, for an abelian group one is not suprised that a small number of small-order generators gives you only a small number of group elements, because order of composing the generators does not matter. But $D_n$ and $S_n$ are both nonabelian beyond the very smallest cases, yet they behave very differently.)