Product of all normal subgroups of a group

Question

If a group has normal subgroups then is it correct that the product of all the normal subgroups of a group is equal to the group itself? I am doing the exercises of the direct product so the question came through my mind that could this be true or not and if that is not true then what are some examples that does not follow?

Answer 1

This is most easily seen to be false with Abelian groups, since all subgroups of Abelian groups are normal. Consider $G = \mathbb{Z} / 4\mathbb{Z}$. You can check that $\mathbb{Z} / 2\mathbb{Z}$ is the only non-trivial, proper, normal subgroup of $G$. It is certainly not the case that $G \cong \mathbb{Z} / 2\mathbb{Z}$. It's not even the case that $G \cong \mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$.

Answer 2

$G$ is a normal subgroup of itself, hence by that the claim is trivially true. If we exclude $G$ itself, the claim is often false. For example if $G=S_5$, then the only non-trivial proper normal subgroup is the simple group $A_n$

Answer 3

There are a lot of counterexamples.

Abelian group $\mathbb{Z}_4$, which is given by Charles Hudgins.

Non-abelian solvable group $A_4$: the only non-trivial proper normal subgroup is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.

Non-abelian unsolvable group $S_n$ for $n\ge 5$: the only non-trivial proper normal subgroup is $A_n$.

But one example that your statement valids is $Q_8$, each of whose subgroup is normal.