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Let $G$ be a group having finite normal subgroups $N_1,... , N_k$ , such that the gcd of $|N_i |$ and $|N_j |$ is $1$ whenever $i\neq j$. Show that $N_1N_2 \cdots N_k \cong N_1\times N_2\times\cdots\times N_k$.

I think one way to solve this exercise is by proving two things: (1) $N_1N_2 \cdots N_k=G$ and (2) $(N_1\cdots N_s)\cap N_{s+1}=\{e\}$. Since in this way we can conclude that $G$ is the internal direct product of $N_1,... , N_k$ and therefore isomorphic to $N_1\times\cdots\times N_k$. The problem is, I don't know exactly how to start. Any hint?


Edition

For $(2)$, let $N=N_1N_2\cdots N_{k-1}$. We claim that $N\cap N_{k}=\{e\}$. Suppose otherwise, that $e\neq a\in N\cap N_{k} $, then $|a|$ must divide the order of both subgroups $N$ and $N_k$, but $(|N|,|N_k|)=1$; thus, $|a|=1$ and so $a=e$. For $(1)$, since $N\cap N_{k}=\{e\}$, we know that $|NN_k|=|N||N_k|\leq |G|$. Here is my problem, how can I prove that Indeed $|N||N_k|= |G|$?

I have not been able to advance much with the Grossmann's hint

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3 Answers 3

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Hint: Because $N_k$ is normal, the quotient map $N_1\cdots N_{k-1}N_k \to (N_1 \cdots N_{k-1}N_k)/N_k \cong N_1 \cdots N_{k-1}$ is a homomorphism.

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  • $\begingroup$ I'm sorry Mr. Grossmann. I've been thinking about it, but I don't get it. $\endgroup$
    – Hopmaths
    Mar 2, 2021 at 23:42
  • $\begingroup$ @Hopmaths A further hint: construct an isomorphism from $N_1 \cdots N_{k-1}N_k$ to $(N_1 \cdots N_{k-1}) \times N_k$ and proceed inductively. $\endgroup$ Mar 3, 2021 at 0:28
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By induction, it is enough to show this when $k=2$. First of all, for $n_1\in N_1$ and $n_2\in N_2$, we have $n_1n_2=n_2n_1$, since you can show that $n_1n_2n_1^{-1}n_2^{-1}\in N_1\cap N_2=1$.

Thus, the map $N_1\times N_2\to N_1N_2\subseteq G:(n_1,n_2)\mapsto n_1n_2$ is a group homomorphism. Now, all that is left is to show injectivity and surjectivity.

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  • $\begingroup$ What? I feel very silly :D. The problem never actually asked me that $G$ be isomorphic to $N_1\times \cdots \times N_k$. That's why I was in a hole. Thank you $\endgroup$
    – Hopmaths
    Jun 2, 2021 at 18:53
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Let $H$ and $K$ be subgroups of $G$. Then,

$\textbf{Lemma 1}$: $H \cap K$ is a subgroup of $G$.
Proof: For $a, b \in H \cap K$, $ab^{-1} \in H$ and $ab^{-1} \in K$ $\implies ab^{-1} \in H \cap K$ $\implies H \cap K \subseteq G$.
Corollary: $H\cap K$ is a subgroup of $H$ and $K$.

$\textbf{Lemma 2}$ (Well-known): Let $H, K$ be subgroups of $G$. Then, $HK$ is a subgroup of $G$ $\iff$ $HK=KH$. Proof: For instance, refer here

$\textbf{Lemma 3}$(Well-known): Let $H, K$ be subgroups of $G$. Then, $|HK|$ $= \dfrac{|H||K|}{|H\cap K|}$.
Proof: For instance, refer here

Let $N_{ij} = N_i \cap N_j$. From Corollary of Lemma 1, $N_{ij}$ is a subgroup of $N_i, N_j$. Therefore, by Lagrange's theorem, $|N_{ij}|$ divides $|N_i|$ and $|N_j|$. Since $(|N_i|,|N_j|)=1$, $|N_{ij}| = 1$. Therefore, $N_{ij} = \{e\}$, where $e$ is the identity of $G$.

Now, $N_iN_j = \{n_in_j|n_i \in N_i, n_j\in N_j\}$. Because $N_i$ are normal, $n_iN_j = N_jn_i$ $\forall n_i \in N_i$. Therefore, $\forall n_i \in N_i$, $\forall n_j \in N_j$, $\exists n_j'\in N_j$ so that $n_in_j=n_j'n_i$ $\implies N_iN_j\subseteq N_jN_i$. Similarly, $N_jN_i\subseteq N_iN_j$ $\implies$ $N_iN_j=N_jN_i$. Therefore, Lemma 2 implies $N_iN_j$ is a subgroup of $G$. Clearly, $N_iN_j$ is normal, as $gN_iN_j$ $=N_igN_j$ $= N_iN_jg$ $\forall g\in G$. Also, from Lemma 3, $|N_iN_j|=|N_i||N_j|$. So, $(|N_iN_j|,|N_{\ell}|)=1$. By induction, $N_1N_2\dots N_k$ is a normal subgroup of $G$, and $(|\Pi_{i=1, i\neq j}^k N_i|, |N_j|) = 1$.

Notice that $N_i \subseteq N_1N_2\dots N_k$ because $e\in N_j \forall j$. By definition of internal direct product, it is now direct that $$N_1N_2\dots N_k = N_1\times N_2 \times \dots\times N_k$$ $\blacksquare$.

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