All groups of order $pqr$ are embedded in $S_{p+q+r}$?

Question

This question was asked in TIFR GS-1 Mathematics test 2025.

State whether the statement is true or false:

Let $m$ be a positive integer, and $S_m$ the symmetric group on $m$ letters. Let $n$ be a positive integer such that for every group $G$ of order $n$, there exists an injective group homomorphism $G \hookrightarrow S_m$. Then $m \geq n$.

I know this is false and we have a simple counterexample: For every group $G$ of order $6$, there exists an injective group homomorphism $G\hookrightarrow S_5$ and $5<6$.
Explanation: There are only two groups of order $6$ up to isomorphism: $S_3$ and $\mathbb Z_6$. They're both embedded in $S_5$.

I want to create a class of counter examples, so I thought of this:

Let $p$ and $q$ be distinct primes. Show that every group of order $pq$ is embedded in $S_{p+q}$.

Here's my attempt at the proof:

WLOG, suppose $p<q$. I know that there are at most two groups of order $pq$ up to isomorphism: $\mathbb Z_{pq}$; and if $p\mid (q-1)$ then $\mathbb Z_p \ltimes \mathbb Z_q$.
It's easy to embed $\mathbb Z_{pq}$ in $S_{p+q}$ as the cyclic group generated by the permutation $(1,\ldots, p)(p+1,\ldots, q)$.
Assume $p\mid (q-1)$, now we show that $\mathbb Z_p \ltimes \mathbb Z_q$ embeds in $S_{p+q}$. The normalizer $G$ of the cyclic subgroup $H=\langle (12\ldots q)\rangle$ in $S_q$ is of order $q(q-1)$. Now $G$ acts on $H$ via conjugation. Since $p\mid (q-1)$, $G$ admits a subgroup $K$ of order $p$. Now, by restriction of the action of $G$, we see that $K$ also acts on $H$. This action is non-trivial since $H$ is self-centralizing. Thus, $K\ltimes H\leq G\leq S_q\leq S_{p+q}$. $\blacksquare$

I wonder if this is true:

Let $p$, $q$ and $r$ be distinct primes. Every group of order $pqr$ in embedded in $S_{p+q+r}$.

Or any further possible generalization?

Answer 1

Let $G$ be a group whose order is square-free, that is, we have $|G|=p_1p_2 \cdot\ldots\cdot p_n$ for distinct primes $p_i$. Then there exists an injective group homomorphism $G \hookrightarrow S_{p_1+p_2+\ldots+p_n}$

Proof: Any group of square-free order is solvable. By Hall's theorem, we may find a collection of subgroups $H_1, H_2, \ldots, H_n\leq G$ such that for all $i$, $H_i$ has index $p_i$ in $G$.

Now consider the set $X=\bigsqcup_i G/H_i$. As $G$ acts by left multiplication on $G/H_i$, it also acts on $X$. By construction $|X|=p_1+p_2+\ldots+p_n$. If we denote the kernel of the associated homomorphism to $S_{p_1+p_2+\ldots+p_n}$ by $K$, then we get $$K \leq \bigcap_i H_i$$

This intersection is trivial, because by Lagrange we get that $|\bigcap_i H_i|$ divides $\gcd_i |H_i| =\gcd_i \frac{p_1p_2 \ldots p_n}{p_i}=1$.

Thus the associated homomorphism from $G$ to $S_{p_1+p_2+\ldots+p_n}$ is injective.