# Is $H_nH_m = G$ if $|G|=nm$ and $H_n, H_m$ are the only subgroups of $G$ of order $n, m$ respectively?

I want to use this and the fact that $$H_n \cap H_m = \{1\}$$ which I've already proved to show that $$G \cong H_n \times H_m$$

Since this are subgroups, and are the only subgroups like this, $$H_n$$ and $$H_m$$ are normal. Then $$H_n H_m \triangleleft G$$. Then, I show that $$|H_n H_m| \geq nm$$ by saying the following:

I have $$n$$ possible elections of $$h_n \in H_n$$ and $$m$$ possible elections of $$h_m \in H_m$$. Then if $$|H_n H_m| < nm$$, then it should be some $$h_{n_1}, h_{n_2} \in H_n$$ and $$h_{m_1}, h_{m_2} \in H_m$$ such that $$h_{n_1}h_{m_1} = h_{n_2}h_{m_2}$$, but this cannot be because $$H_n \cap H_m = \{1\}$$. Then $$|H_nH_m| \geq nm$$. Trivially, $$|H_nH_m| \leq nm$$, then $$|H_nH_m|=nm$$ so $$G = H_nH_m$$.

Is this correct?

• Is the fact that $H_n\cap H_m=\{1\}$ given? I don't think it can be proven just from the information in your title : for example the cyclic group of order $n^3$ contains only one subgroup of order $n$ and one of order $n^2$, but they are not disjoint. May 10, 2019 at 16:30
• In general, for subgroup $H$ and $K$, we have that $|HK| |H\cap K| = |H||K|$ (in the sense of cardinality). This, regardless of whether $HK$ is a subgroup or not. If $|H\cap K| = 1$, and $|H||K|=|G|$ is finite, then it follows that $|HK|=|G|$ and since $HK\subseteq G$, we get $HK=G$. However, it need not be a direct product; e.g., $G=S_3$, $H=\{\mathrm{id}, (123), (132)\$ and $K=\{\mathrm{id},(12)\$. May 10, 2019 at 18:54

Consider the function $$\phi: H_n \times H_m \to G$$ given by $$(u,v) \mapsto uv$$.
Then $$\phi$$ is injective because $$H_n \cap H_m = \{1\}$$. Indeed, $$u_1 v_1 = u_2 v_2$$ implies $$u_2^{-1} u_1 = v_2 v_1^{-1} \in H_n \cap H_m = \{1\}$$.
Therefore, $$\phi$$ is surjective because both sets have the same size, $$mn$$ elements.
Since the image of $$\phi$$ is $$H_n H_m$$, it is equal to $$G$$.
• I think I put the question slightly wrong. I've edited it just now. What I tried to prove there is that $H_nH_m = G$. May 10, 2019 at 16:20