# Show that $N_1N_2 \cdots N_k \cong N_1\times N_2\times\cdots\times N_k$ .

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

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.

• I'm sorry Mr. Grossmann. I've been thinking about it, but I don't get it. Mar 2, 2021 at 23:42
• @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. Mar 3, 2021 at 0:28

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.

• 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 Jun 2, 2021 at 18:53

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 $$N_1N_2\dots N_k = N_1\times N_2 \times \dots\times N_k$$ $$\blacksquare$$.