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I have a problem. It states that:

Let $G$ is a group and $|G|=mn$, $(m,n)=1$. Assume that $G$ has exactly one subgroup $M$ with order $m$ and one subgroup $N$ with order $n$. Prove: $G$ is a direct product of $M$ and $N$.

Here is my approach:

Obviously, we have $M\cap N = 1$. By Product Formula we have: $|MN|=|M|\cdot|N|=mn$.

Let $m_1$, $m_2$ be in $M$ and $n_1$, $n_2$ in $N$. If $m_1n_1=m_2n_2$ then $m_2^{-1}m_1=n_2n_1^{-1}$. By $M\cap N = 1$ we must have $m_1=m_2$, $n_1=n_2$. So those elements in the form $m_in_j$ with $m_i\in M$, $n_j\in N$ are distinct. Because $|G|=mn$ we must have $G=MN$.

I'm kind of confused because it seems the uniqueness of subgroups $M$ and $N$ is useless. Is my proof still correct without this hypothesis or I mislead at certain point? Please explain to me.

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In general, $MN=\{mn\mid m\in M, n\in N\}$ is not a group, but in this case $MN$ is a group by considering its cardinality as comment said. But $G=MN$ does not implies $G\simeq M\times N$. To show that $G$ is isomorphic to $M\times N$, we have to show that elements of $M$ and $N$ commute, i.e. $mn=nm$ for all $m\in M, n\in N$. By the uniqueness condition we can show that $M, N$ are normal subgroups of $G$. Then $mnm^{-1}n^{-1}\in M\cap N$ (why?), so $mnm^{-1}n^{-1}=e$ and they commute. Then $(m,n)\to mn$ became an isomorphism between $M\times N$ and $MN=G$.

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    $\begingroup$ In this case $MN$ is a group because it is equal to all of $G$ by considering cardinality: $|MN| = |M| |N| / |M \cap N| = |G|$. $\endgroup$ – Gregory Simon Jul 25 '16 at 5:34
  • $\begingroup$ @GregorySimon I edited. Thanks! $\endgroup$ – Seewoo Lee Jul 25 '16 at 5:46
  • $\begingroup$ Yeah. According to Intro Group Theory by Rotman Product Formula still true even $M$.$N$ is not group. And by compare the the number of elements we must have $M.N$=$|G|$. I really want to understand the motivate behind the hypothesis, why he puts it in this context ? $\endgroup$ – Anh_Rose 1210 Jul 25 '16 at 5:46
  • $\begingroup$ @AnhVũ I edited answer few seconds ago. Please check it. $\endgroup$ – Seewoo Lee Jul 25 '16 at 5:48
  • $\begingroup$ Thanks ! It's very helpful. $\endgroup$ – Anh_Rose 1210 Jul 26 '16 at 16:34
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Just because every element in $G$ can be written uniquely as $nm$ that does not mean $G = M\times N$. For example, $G$ could be a semidirect product, like a dihedral group.

By the way, the notation $G=M.N$ usually means an "extension of $N$ by $M$" which means $M$ is (isomorphic to) a normal subgroup of of $G$ and $G/M \cong N$. This is much weaker than you are trying to show.

This is from the Atlas of Finite Groups:

notation for extending groups

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You say "obviously , we have $M\cap N=1$". This "obvious" is not obvious at all. At this point you dont use the hypothesis.

This is generaly how one detect mistakes in proofs : when the author write "obviously", or "it is easy to see that", there is a serious doubt.

To prove the "obvious", just says that $M\cap N$ hase order dividing the order of $M$ and $N$ (Lagrange), so the order of this group is 1.

By the way, if $p$ is a prime number, let $G=S_p$ the symmetric group on $\{1,...p\}$ letters, $M=S_{p-1}$ the symmetric group on $\{1,...p-1\}$ letters and $H$ the subgroup of $G$ generated by a cyclic permutation $(1,2,...p)$. Then $G=MN$ but this is not a product group.

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  • $\begingroup$ This isn't an answer to the question. Worthy of a comment at best. And depending on your level of understanding, it may be obvious $\endgroup$ – Aweygan Jul 25 '16 at 5:55
  • $\begingroup$ @Aweygan. I just pointed out the point where the hypothesis is not used. To my opinion the word "obvious" is a personal impression that should never appears in a mathematical proof. The argument involve Lagrange theorem, so considering the level of the question, an important not obvious fact. $\endgroup$ – Thomas Jul 25 '16 at 10:30
  • $\begingroup$ Your example also fails the OP's hypothesis though, as neither $M$ nor $H$ is the unique subgroup of their order. $\endgroup$ – Alex Meiburg Aug 29 '16 at 8:21

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