Let $G$ be a finite abelian group of order $pq$, where $p<q$ are both primes. I want to show that $G$ is isomorphic to a subgroup of $S_{p+q}$ (but is not isomorphic to any subgroup of order $S_{p+q-1}$).
What I know so far, assuming $|G|=pq$, $p<q$: $G$ is abelian iff there is only one $p$-subgroup and in this case $G\cong \mathbb{Z}/pq\mathbb{Z}$. I also know that if $G$ is not abelian, then $G$ is isomorphic to a subgroup of $S_q$ but not to a subgroup of $S_{q-1}$ but I don't think this helps
I'm trying to find a set $C$ with $p+q$ elements such that there is a faithful action/representation $\rho: G\rightarrow B(C)$, where $B$ is the group of bijections $C\rightarrow C$, but without success.
Does anybody have any idea?
$G$ is a product of two cyclic groups. There is a natural way to embed into $S_p \times S_q \le S_{p+q}$.
For $S_{p+q-1}$, $G$ has an element of order $pq$. The order of a permutation is the lcm of its cycle lengths. So the corresponding permutation must have a cycle of length $p$ and another of length $q$, whence it cannot belong to $S_{p+q-1}$.