Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.8
Prove that if $n \neq m$ then $S_n$ and $S_m$ are not isomorphic.
Solution: We know that $|S_n| = n!$ and $|S_m| = m!$, but if $n \neq m$ then $n! \neq m!$.
Finite symmetric groups of different orders are not isomorphic
