A group $G$ can be shown to be the direct product of 2 of its normal subgroups if they generate the entire group and intersect trivially. Does this extend to more than 2 subgroups? For example, if $G$ has 3 normal subgroups that generate the entire group and pairwise intersect trivially, is $G$ necessarily isomorphic to their direct product?


Suppose that $G=A_{1}...A_{n}$ where $A_{i} \lhd G \;\forall 1 \leq i \leq n$. Let $A_{i}':=A_{1}...A_{i-1}A_{i+1}...A_{n}$. Note that $A_{i}' \lhd G$ by the result that if $A,B \lhd G$, then $AB \lhd G$. Suppose also that $A_{i}' \cap A_{i} = \{1\} \; \forall i$. It can then be shown (inductively) that $G \cong A_{1} \times ... \times A_{n}$.


No, but you are close.

$C_2 \times C_2$ has three normal subgroups that generate the entire group and pairwise intersect trivially: $C_2 \times 1$, $1\times C_2$, and $\Delta = \{ (x,x) : x \in C_2 \}$. Each is isomorphic to $C_2$, but $C_2 \times C_2$ is not isomorphic to $C_2 \times C_2 \times C_2$.

The condition is very similar though, $G=A \times B \times C$ iff (1) $G=ABC$, (2) $A,B,C \unlhd G$, and (3) $AB \cap C = AC \cap B = BC \cap A = 1$.

You got (1) and (2) right, and (3) is close. You just leave out one factor each time.

  • $\begingroup$ Ah, so for higher numbers of factors, you need each subgroup to have trivial intersection with the product of ALL the other subgroups? $\endgroup$ – Nishant Jan 11 '14 at 0:17
  • $\begingroup$ Yup. I think Edward ffitch is writing up that answer. You can think of $A \times (B \times C)$ means $A \cap BC=1$ and $B \cap C = 1$, but many people like symmetric conditions better. $\endgroup$ – Jack Schmidt Jan 11 '14 at 0:20

This notion extends only to finite direct products, which @Edward's answer demonstrates quite well. Below I'll show you what happens when you try to do this with infinite direct products (countable or otherwise).

Let $G$ be a group and $\{G_\alpha\}_{\alpha\in A}$ be a family of normal subgroups of $G$ with the property that $G_\alpha\cap \langle G_\beta \,|\, \beta \in A\setminus \{\alpha\} \rangle = 1$ for all $\alpha$ and $G=\langle G_\alpha \,|\,\alpha \in A\rangle$. Let $G^\star = \prod_{\alpha \in A}G_\alpha$ be the standard (external) direct product of these subgroups. Any $g\in G$ may be uniquely represented in the form $g=g_{\alpha_1}g_{\alpha_1}\cdots g_{\alpha_n}$, where each $g_{\alpha_i}$ are members of $G_{\alpha_i}$, and the $\alpha_i$ are distinct.. Define $\pi_\alpha:G\rightarrow G_\alpha$ by $\pi_\alpha(g)=g_\alpha$ as in the previous sentence (taking the value of the identity if $g$ does not have an $\alpha$-component). Now define $\pi:G\rightarrow G^\star$ as $\pi(g)=(g_\alpha)_{\alpha\in A}$. We see that $\pi$ is injective (though not in general surjective, as $\prod_\alpha G_\alpha$ can have elements like $(g_\alpha)_{\alpha\in A}$ where all $g_\alpha$ are nontrivial). $\pi(G)$ is therefore isomorphic to the subgroup of $G^\star$ comprised of all elements for which all but finitely many coordinates are equal to the identity; this is the restricted external direct product, notated by $\oplus_\alpha G_\alpha$.

  • $\begingroup$ Is there a reason why $G$ cannot have an element $g$ whose representation in terms of the $G_\alpha$s is infinite? $\endgroup$ – Nishant Jan 11 '14 at 14:46
  • $\begingroup$ @Nishant yes, because we assumed that $\cup_\alpha G_\alpha$ was a generating set for $G$. by definition any element may be written as a finite product of elements from that set. $\endgroup$ – Alexander Gruber Jan 12 '14 at 16:06

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