Why ${S_{n-1}}$ is not a subgroup of ${S_n}$

Question

I know that in general this is not true. Also, I know that subgroups of index n of ${S_n}$ are isomorphic to ${S_{n-1}}$ but why they are not subgroups? For example, why ${S_6}$ is not a subgroup of ${S_7}$? We have closure, existence of inverses and unit element. What do I miss here? Thanx

Answer 1

Just because you can view a group as a subgroup of another (i.e., via an isomorphism with the subgroup) doesn't mean that it really is a subgroup.

For a perhaps more enlightening example, consider $\mathbb{Z}/2$. This is not a subset of $\mathbb{Z}/4$ in the strict sense of the term, but if we see the former as $\{0,1\}$ and the latter as $\{0,1,2,3\}$, it is. However, this doesn't exhibit $\mathbb{Z}/2$ as a subgroup of $\mathbb{Z}/4$. On the other hand, one can see $\mathbb{Z}/2$ as a subgroup of $\mathbb{Z}/4$ by viewing the former as $\{0,2\}$ instead of $\{0,1\}$. So, knowing this, would you say that $\mathbb{Z}/2$ is without a shadow of a doubt a subgroup of $\mathbb{Z}/4$?