# Remark about direct product of groups

This is an excerpt from Herstein's book and I would like to understand the meaning of text which I've marked with red line.

The author says that the theorem proves the following thing: If $G\cong G_1\times G_2\times \dots \times G_n$ then $G=\bar{G_1}\bar{G_2}\dots \bar{G}_n$, where $\bar{G}_i\cong G_i$.

However, after detailed reading the statement and proof of theorem I've realized that theorem and paragraph (marked with red) are somewhat converse facts. Maybe I am wrong.

Notation: By $\bar{G}_i$ Herstein means the following $\{(e_1,\dots,e_{i-1}, g_i, e_{i+1},\dots, e_n): g_i\in G_i\}$ where $e_i$ - the unit in group $G_i$.

Can anyone explain in detail why the theorem proves the red paragraph?

• Well, a more direct way of stating what that paragraph say is: an external direct product of groups is the internal product of subgroup isomorphic to to the factors.. – Mariano Suárez-Álvarez Feb 1 '18 at 18:30
• @MarianoSuárez-Álvarez, I guess that paragraph means the following: $G_1\times G_2\times \dots \times G_n \cong \bar{G}_1\bar{G}_2\dots \bar{G}_n$ where $\bar{G}_i\cong G_i$. – ZFR Feb 1 '18 at 18:35
• Exactly (and where on the right the product is direct) – Mariano Suárez-Álvarez Feb 1 '18 at 18:36
• @MarianoSuárez-Álvarez, Does it follow from theorem? If yes, could you explain it, please? – ZFR Feb 1 '18 at 18:37
• @MarianoSuárez-Álvarez, Maybe i am misunderstanding something :/ – ZFR Feb 1 '18 at 18:39

So you have a chain of isomorphisms $$\Gamma\stackrel{\Psi}{\to} G_1×⋯×G_n\stackrel{\phi}{\to} \bar{G_1}\ldots \bar{G_n}$$

Now set $$\Gamma_i=\Psi^{-1}\left(\phi^{-1}(\bar{G_i})\right)$$

Simply because the two maps are isomorphisms give you everything you need:

• The $\Gamma_i$ are normal subgroups of $\Gamma$
• $h_i\in \Gamma_i$ and $h_j\in \Gamma_j$ will commute as soon as $i\neq j$
• $\Gamma_i\ \cap \left<\bigcup_{j\neq i} \Gamma_j\right>=\left\lbrace e\right\rbrace$

All of these properties are structural (preserved via isomorphisms).

Therefore $\Gamma\cong \Gamma_1 \ldots\Gamma_n$ where $\Gamma_i\cong G_i$ for all $i$.

• But the book states that $\Gamma=\bar{G}_1\bar{G}_2 \dots \bar{G}_n$. However you have not this in your answer. – ZFR Feb 2 '18 at 19:56
• Dear Arnauld, I've aksed the question. Could you answer it, please? – ZFR Feb 3 '18 at 15:48
• Hi, sorry I was away. If you look carefully the red paragraph indicates "isomorphic to an internal product of subgroups that are isomorphic to the $\bar{G_i}$". Not "=", just "isomorphic". – Arnaud Mortier Feb 3 '18 at 17:20