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I want to ask a question, and I found it here: Why is the direct product of a finite number of nilpotent groups nilpotent?

But I am struggling to understand how can we take the product of two normal series and still have a normal series, that is, for the case of two subgroups, if $H\triangleleft K$ and $H'\triangleleft K'$ why can we deduce that $HK\triangleleft H'K'$ ?

Another question is to show the containment in the original post in the link above, many answers show the result differently, it should be possible to show the containment directly no? Using the definition of nilpotence by a central series (not by an upper or descending central series).

NotaChoice
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1 Answers1

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Let me answer your first question.

This is a situation where good notation really helps.

Let me use direct product notation: $K \oplus K'$ instead of $KK'$, and similarly $H \oplus H' < K \oplus K'$ instead of $HH' < KK'$.

Also, let me express each element of $K \oplus K'$ as an ordered pair $(k,k')$ with $k \in K$ and $k' \in K'$, and with group operations given coordinate wise, namely $$(k_1,k'_1) (k_2,k'_2) = (k_1 k_2, k'_1 k'_2) \qquad (k,k')^{-1} = (k^{-1},k'{}^{-1}) $$ So now take any $(h,h') \in H \oplus H'$ and any $(k,k') \in K \oplus K'$ and let's calculate: $$(k,k')^{-1} (h,h') (k,k') = (khk^{-1},k'h'k'^{-1}) $$ From the assumption that $H$ is normal in $K$ and $H'$ is normal in $K'$ it follows that $khk^{-1} \in H$ and $k'h'k'{}^{-1} \in H'$ and therefore $(khk^{-1},k'h'k'^{-1}) \in H \oplus H'$. This proves that $H \oplus H'$ is normal in $K \oplus K'$.

Lee Mosher
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    Your group operation seems to assume that all elements of $K$ commute with all elements of $K'$, but that is not true in general. – Derek Holt Jan 25 '24 at 09:35
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    Hmm... do you mean you are taking the 2nd paragraph of the proof to not assume a direct product? – Lee Mosher Jan 25 '24 at 13:52
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    On re-reading, I see that the second paragraph is very unclear. It talks about "products" rather than "direct products" and it appeared to be an exact duplicate of the earlier question, which was closed, where $H$ and $H'$ were just normal subgroups of an arbitrary group $G$. In any case I think the question deserves to remain closed until the OP clarifies. – Derek Holt Jan 25 '24 at 14:05
  • Sounds appropriate. – Lee Mosher Jan 25 '24 at 14:23
  • Can I ask you to explain what you mean by just a product not a direct product? But this question needs to be deleted. – NotaChoice Jan 26 '24 at 20:44