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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Arnaud D.
    May 10, 2019 at 16:30
  • 1
    $\begingroup$ 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)\$. $\endgroup$ May 10, 2019 at 18:54

1 Answer 1


Your argument is correct but can be written more clearly as follows.

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 $G$.

  • $\begingroup$ I think I put the question slightly wrong. I've edited it just now. What I tried to prove there is that $H_nH_m = G$. $\endgroup$
    – Silkking
    May 10, 2019 at 16:20

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