# A finite group of order $mn$ with $m,n$ relatively prime, together with subgroups of orders $m, n$.

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 $$N$$.

• 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 $6.1$. Jan 1 at 11:35
• 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? Jan 1 at 11:41

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 $$G\rightarrow G/A\times G/B\rightarrow S_n\times S_m\rightarrow S_{m+n}$$