Let $G$ be a finite group of order $mn$ with $(m,n) = 1$. Assume that there exist subgroups $M,N$ of $G$ of orders $m$ and $n$, respectively. Prove that $G$ is isomorphic to a subgroup of the symmetric group $S_{m+n}$.

It is easy to notice that $M\cap N = \{e\}$ since $m,n$ relatively prime, thus, $|MN|=\frac{|M||N|}{|M\cap N|} = mn$. Hence, $G = MN = NM$.

To prove that $G$ is isomorphic to a subgroup of the symmetric group $S_{m+n}$, I think I should construct a group action (G acting on some set of order m+n). And the set I think of is $A = M \cup N$ where identities in $M$ and $N$ respectively should be viewed as two different elements. Hence $|A| = m+n$. If elements in $M$ commute with those in $N$, I can define a group action $hk \cdot a = ha$ if $a \in M$ or $ka$ if $a \in N$. (Here, $h \in M, k \in N$, and $hk$ represents an element in $G$). However, I do not know how to prove the case where elements in $M$ do not commute with those in $N$.

  • $\begingroup$ See the Minimal permutation degree of finite groups. If, say, $G=M\rtimes N$, then $d(G)\le |M|+d(N)=m+n$, see Lemma $6.1$. $\endgroup$ Jan 1 at 11:35
  • $\begingroup$ This might require either $M$ or $N$ to be normal, but this is not given, not sure I can deduce it from the conditions given? $\endgroup$
    – Maksim
    Jan 1 at 11:41

1 Answer 1


Basic steps.

  1. Let $G/M$ the collection of left cosets of $M$ in $G$, $n=|G:M|$. Then there exists a natural group action of $G$ on $G/M$ by left multiplication. Let $A$ be the kernel of the homomorphism $G\to S_n$ associated with this action. We have $A\leq M$.

  2. Similarly, we have an embedding $G/B\rightarrow S_m$ and $B\leq N$.

  3. Since $A\cap B=1$ we have an injective homomorphism $$G\rightarrow G/A\times G/B\rightarrow S_n\times S_m\rightarrow S_{m+n}$$


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