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In my book, I have a theorem that says the following:

Let $G$ be a group. If $G_1,G_2$ are subgroups such that:

  • $G_1, G_2 \lhd G$
  • $G_1G_2 = G$
  • $G_1 \cap G_2 =\{e_G\}$

Then $G \cong G_1 \times G_2$

Later, there is a remark that says that the converse of the theorem also holds. So, I suppose this means, that if we have that $G \cong G_1 \times G_2$ for subgroups $G_1,G_2$, then the three conditions listed above hold.

The 'proof' goes as follows:

If $G = G_1 \times G_2$, then $G = H_1H_2 $ with $H_1 = G_1 \times \{e_{G_2}\}$ and $H_2 = \{e_{G_1}\} \times G_2$. The groups $H_1,H_2$ are normal in $G$ and $H_1 \cap H_2 = \{e\} \quad \triangle$

Can someone explain this please? I really don't get how this proves anything. They don't even start from $G \cong G_1 \times G_2$? I will award the bounty to the person who can give me a detailled and rigorous explanation.

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  • $\begingroup$ They do start from $G \cong G_1 \times G_2$! Their $H_1$ is defined as $\{ (g_1, e_{G_2} ) : g_1 \in G_1 \} \subset G_1 \times G_2$, and their $H_2$ is defined similarly. It then remains to verify your three bullet points... $\endgroup$ – Kenny Wong Jun 23 '17 at 14:09
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    $\begingroup$ One must be careful with the converse. Let $G = \mathbb{Z}^{\mathbb{N}}$, $G_1 = \{x \in G : x_0 = 0\}$, $G_2 = \{ x\in G : x_0 = x_1 = 0\}$. Then $G \cong G_1 \times G_2$, but $G_1G_2 \neq G$ and $G_1 \cap G_2 \neq \{e\}$. $\endgroup$ – Daniel Fischer Jun 23 '17 at 14:21
  • $\begingroup$ Isn't this a counterexample then? $\endgroup$ – user370967 Jun 23 '17 at 14:32
  • $\begingroup$ If you look at the proof, it doesn't start from $G\simeq G_1 \times G_2$, but from $G=G_1\times G_2$. $\endgroup$ – Max Jun 23 '17 at 17:05
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    $\begingroup$ I think your problems are coming from a sloppy formulation of the converse. The only version that makes sense is: if $G\cong G_1\times G_2$, then there are subgroups $H_{1,2}\subseteq G$, isomorphic to $G_{1,2}$, such that etc. etc. (Otherwise, what stops you from for example taking $G_1=G_2$, when the third bullet point can't hold.) With this obvious adjustment, the difference between $=$ and $\cong$ has actually disappeared now. $\endgroup$ – user138530 Sep 20 '17 at 6:20
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Definition. The cartesian product of two groups $G=G_1\times G_2$ is a group under the operation $(a,b)(c,d)=(ac,bd)$ together with these two maps: $$ \begin{equation} p_1\colon G\to G_1;~(x_1,x_2)p_1=x_1\\ p_2\colon G\to G_2;~(x_1,x_2)p_2=x_2 \end{equation} $$

(Theorem. $p_1$ and $p_2$ are group homomorphisms, and $\forall x_1\in G_1,\forall x_2\in G_2,\exists|x\in G$ such that $xp_1=x_1$ and $xp_2=x_2$. Moreover let $$ \begin{equation} i_1\colon G_1\to G;~x_1i_1=(x_1,e_{G_2})\\ i_2\colon G_2\to G;~x_2i_2=(e_{G_1},x_2) \end{equation} $$ it results that $$i_np_m=\begin{cases}I_{G_n}&\text{ if }n=m\\ O_{G_nG_m}&\text{ if }n\ne m \end{cases}$$ where $n,m\in\{1,2\}$ and $O_{XY}\colon X\to Y;~x\mapsto e_Y$ with $Y$ group. )

Definition. A direct product of two groups usually written again as $G_1\times G_2$ but that we will write as $A=(\{G_1, G_2\}, \{q_1,q_2\})$ is a group under the operation $(a,b)(c,d)=(ac,bd)$ together with these two homomorphisms: $$ \begin{equation} q_1\colon A\to G_1\\ q_2\colon A\to G_2 \end{equation} $$ for which $\exists$ homomorphisms $j_k\colon G_k\to G$ where $k=1,2$ such that $$j_nq_m=\begin{cases}I_{G_n}&\text{ if }n=m\\ O_{G_nG_m}&\text{ if }n\ne m \end{cases}$$ where $n,m\in\{1,2\}$

Definition. A direct-product group isomorphism is a group isomorphism $T:A\to B$, where $A=(\{G_1,G_2\},\{q_1,q_2\})$ and $B=(\{G_1,G_2\},\{r_1,r_2\})$ direct products of two groups $G_1$ and $G_2$, such that $$ \begin{equation} r_m=Tq_m \end{equation} $$ where $m=1,2$

(Theorem. The cartesian product is a direct product, that is, $G_1\times G_2=(\{G_1, G_2\}, \{p_1, p_2\})$, where $p_1$ and $p_2$ have been defined in the first of our definitions above. Moreover any direct product $A$ of two groups $G_1$ and $G_2$ is direct-product group isomorphic to the cartesian product group $G_1\times G_2$)

All that said, what you name 'proof' is not the proof of the converse of the main theorem but a lemma, probably made difficult to read because the letter $G$ has been reused with a different meaning w.r.t. that used in the main theorem. So first let's rewrite it avoiding such problem:

Lemma. The cartesian product $G_1 \times G_2$ is such that $G_1 \times G_2 = H_1H_2 $ with $H_1 = G_1 \times \{e_{G_2}\}$ and $H_2 = \{e_{G_1}\} \times G_2$. The groups $H_1,H_2$ are normal in $G_1 \times G_2$ and $H_1 \cap H_2 = \{e\} \quad \triangle$

Now by the main theorem and this lemma it follows:

Corollary. The cartesian product of two groups $G_1\times G_2$ is isomorphic to a direct product of $H_1$ and $H_2$ where $H_1 = G_1 \times \{e_{G_2}\}$ and $H_2 = \{e_{G_1}\} \times G_2$, and in particular $G_1\times G_2\cong H_1\times H_2$, that is $H_1\times H_2=(\{G_1,G_2\},\{q_1, q_2\})$ where $$ \begin{equation} q_1\colon H_1\times H_2\to G_1;~(x_1,0,0,x_2)\mapsto x_1\\ q_2\colon H_1\times H_2\to G_2;~(x_1,0,0,x_2)\mapsto x_2 \end{equation} $$ and $\exists$ an isomorphism $T\colon H_1\times H_2\to G_1\times G_2$ such that $q_n=Tp_n$, where $n=1,2$, indeed $$T\colon (x_1,0,0,x_2)\mapsto(x_1,x_2)$$

Moreover:

  1. being $H_1\lhd G_1\times G_2$ it results that $G_1=H_1T^{-1} \lhd (G_1\times G_2)T^{-1}=H_1\times H_2$
  2. being $H_1H_2=G_1\times G_2$ it results that $G_1G_2=(H_1T^{-1})(H_2T^{-1})=(H_1H_2)T^{-1}=(G_1\times G_2)T^{-1}=H_1\times H_2$
  3. being $H_1\cap H_2=\{e\}$, it results that $G_1\cap G_2=H_1T^{-1}\cap H_2T^{-1}=(H_1\cap H_2)T^{-1}=\{e\}T^{-1}=\{e\}$

Proof of the converse of the main theorem. Let $G$ be a direct product of two subgroups of its, $G=(\{G_1, G_2\}, \{r_1, r_2\})$. Then $G$ is direct-product group isomorphic to the cartesian product of those subgroups $G\cong G_1\times G_2$. But then by the corollary it is also $G\cong H_1\times H_2$. That means that $\exists U\colon G\to H_1\times H_2$ direct-product group isomorphism such that $r_n=Uq_n$, where $n=1,2$. For the corollary $H_1\times H_2$ has two subgroups that satisfy the conditions of the main theorem, but then thanks to $U$, $G$ have two subgroups satisfying those conditions as well.

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  • $\begingroup$ Thanks so much. Finally someone who gives an answer! $\endgroup$ – user370967 Sep 25 '17 at 17:48

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