The following part of Gallian Text ($7^{th}$ Ed., Pg.: $188$)

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suggests that the following result holds:

If for $H_1,H_2,...,H_n\le G$ such that $G\simeq H_1\oplus H_2\oplus...\oplus H_n$ (external direct product) then $G=H_1\times H_2\times...\times H_n$ (internal direct product) which I would like to prove.

That's I've to show that,

$(i) ~H_i\lhd G~\forall~i,~(ii) ~G=H_1H_2...H_n,~ (iii) ~H_1...H_i\cap H_{i+1}=(e)~\forall~i=1,2,...,n-1$

I've got stuck in proving all the above three. Please help.

  • $\begingroup$ Without the finiteness condition, the result fails in general even for the case $n = 1$. $\endgroup$ – Sangchul Lee Sep 18 '13 at 12:54
  • $\begingroup$ The suggested statement is badly formulated. I think what @sos440 wants to say that if $H_1\leq G$ such that $G\cong H_1$ then it does not follow that $G=H_1$. This is indeed obvious, even for $G=\Bbb Z$. $\endgroup$ – Marc van Leeuwen Sep 18 '13 at 13:01
  • $\begingroup$ @MarcvanLeeuwen: I agree with you (since $\Bbb Z\cong\langle2\rangle$ but clearly $\Bbb Z\neq\langle2\rangle$. Is the author's notation to blame for some of the confusion? You can see he writes $G=HK$ and $G=H\times K$, which need not be the same (although they can certainly be isomorphic). $\endgroup$ – Clayton Sep 18 '13 at 13:08
  • $\begingroup$ The statement you should want to prove is that if $H_1,H_2,\ldots,H_n\le G$ and the map on the external product $H_1\oplus\cdots\oplus H_n\to G$ given by $(h_1,h_2,\ldots,h_n)\mapsto h_1h_2\ldots h_n$ is an isomorphism of groups, then $G$ is the internal direct product of its subgroups $H_i$. The notation is very confusing, since the underlying set of the external direct product is the Cartesian product $H_1\times\cdots\times H_n$, not to be confused of course with the internal direct product (which may would write $H_1\oplus\cdots\oplus H_n$, sigh). $\endgroup$ – Marc van Leeuwen Sep 18 '13 at 13:15

Consider in $G$ subgroups $\bar{H}_1=H_1\times\{e\}\times\ldots\times\{e\}$, $\bar{H}_2=\{e\}\times H_2\times\ldots\times\{e\}$ etc. Then prove that $\bar{H}_i\cong H_i$ and $\bar{H}_i$ yield your conditions (i)-(iii).

  • $\begingroup$ One can take all the $H_i$ equal, and G is isomorphic to their external direct product, but G is not an internal direct product unless G=H. $\endgroup$ – Yassine Guerboussa Sep 18 '13 at 13:15
  • $\begingroup$ I think in general, $G$ would be the internal direct product of the inverse images of the $H_i$ by the existing isomorphism. $\endgroup$ – Yassine Guerboussa Sep 18 '13 at 13:19
  • $\begingroup$ @Yassine Guerboussa: Sorry, I didn't understand your remarks.:-( $\endgroup$ – Boris Novikov Sep 18 '13 at 13:23
  • $\begingroup$ Your answer is true, if one requires that $G$ is an internal direct product of $\bar{H_i}$, but the question asks about proving this for the $H_i$ themselves. $\endgroup$ – Yassine Guerboussa Sep 18 '13 at 13:28
  • $\begingroup$ @Yassine Guerboussa: Of course, $G$ is not equal to the internal direct product of $H$. One should to prove either that $G$ is isomorphic to the internal direct product of $H_i$, or that $G$ is the internal direct product of subgroups which are isomorphic to $H_i$. I believe that OP understand this. $\endgroup$ – Boris Novikov Sep 18 '13 at 15:19

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