# G does not have to be isomorphic to direct product of trivially intersecting normal subgroups of G

From Dan Saracino, abstract algebra (Exercise 14.8)

Let $$G_1,...,G_n$$ be subgroups of G such that:

i) $$G_1,...,G_n$$ are all normal;

ii) $$G = G_1.G_2...G_n$$, that is, every element of G can be written as $$g_l.g_2...g_n$$, with $$g_i \in G_i$$;

iii) for $$1\leq i\leq n$$, $$G_i\cap G_1 G_2 . . G_{i-1}=\{e\}$$.

Show that $$G\cong G_1 \times G_2 \times \dots \times G_n$$

I have proved the above, but I am not able to find a counterexample for the below statement.

Show, by an example, that if we replace (iii) by the weaker condition $$G_i\cap G_j = \{e\}$$ for $$i \neq j$$, then G does not have to be isomorphic to $$\cong G_1 \times G_2 \times \dots \times G_n$$

• Use $\times$ for $\times$ and $\{ x\}$ for $\{ x\}$. – Shaun Oct 22 '20 at 20:47
• Well there is an example in which $G$ has order $4$. – Derek Holt Oct 22 '20 at 22:06

The group $$C_2\times C_2$$ contains three proper nontrivial subgroups, any two of which intersect trivially. But the direct product of all of them has order $$8$$.