Let $G\simeq N_1\times N_2$ while $N_i$ are normal subgroups of $G$.then what can we say about those $N_i$?

I'm studying internal direct products of groups. and I've seen that an assumption $N_i \cap N_j=\langle e \rangle$ is always assumed when statement involves something like $G\simeq \prod^w N_i$. So I want to know the sufficient conditions for being $G\simeq \prod^w N_i$.

Let $G\simeq N_1 \times N_2$ while $N_i$ are normal subgroup of $G$.
Is $N_1 \cap N_2 = \langle e \rangle$ or $G=N_1N_2$ hold?
Since $N_1 \times N_2 \simeq N_1N_2$ by an isomorphism $\phi (n_1,n_2)=n_1n_2$, I can easily show $G\simeq N_1N_2$.
but is it follow that $G=N_1N_2$?

This is my first time in group theory. So my question can be quite elementary. Thank you in advance.

• I think there is a result stating that for subgroups $N_1$ and $N_2$ of $G$, the isomorphism $G \simeq N_1 \times N_2$ is true if and only if a) $N_1 \cap N_2 = \{e\}$, b) $N_1 N_2 = G$, and c) $N_1$ and $N_2$ are both normal subgroups of $G$. Aug 3, 2018 at 23:15

I think that the result is not true.

For example, take $G=V=\{e,a,b,ab\}$, the Klein $4$-group, and $N_1=N_2=\langle a\rangle$
Then $G\cong \Bbb{Z}_2\times \Bbb{Z}_2\cong N_1\times N_2$.
But clearly $G\neq N_1N_2$ and $N_1\cap N_2\neq \{e\}$.

In group theory, we usually treat isomorphism already as "equivalence" condition in other subjects. Similar as we will say "$2+3=5$", if we claim that $G\cong N_1N_2$ then we can really treat them as the same group, since all the operations inside are the same (just the change in the naming of the elements. So I guess if you assumed $G\cong N_1N_2$, then in some sense $G$ "$=$" $N_1N_2$ upon changing the naming of elements...

For example, we say $S_3\cong D_3$, but in $S_3$, elements are usually written as $(123)$ or $(23)$ while in $D_3$, they are written as $r$ or $sr$. But if we identify $(123)\mapsto r$ and $(12)\mapsto s$ then they are in some sense "equal".

For other questions you raised, they can be proved. Let $G$ be a group such that $G\cong N_1\times N_2$.

Then for any elements $n_1\in N_1$, they are identified in $G$ by $(n_1,1)$. So let $n_2\in N_2$ be an arbitrary element in $N_2$, we have $(1,n_2)^{-1}\cdot (n_1,1)\cdot (1,n_2) = (1\cdot n_1\cdot 1,n_2^{-1}\cdot n_2 = (n_1,1)$. So $N_1\cong N_1\times \{1\}$ and $N_2\cong \{1\}\times N_2$ are contained in the center of the group, or at least, they are normal groups.

If an element $(p,q)\in G$ are in both $N_1$ and $N_2$, then $(p,q)=(n_1,1)$ for some $n_1\in N_1$, and $(p,q)=(1,n_2)$ for some $n_2\in N_2$. Hence, we can conclude that $(p,q)=(1,1)$ and so $N_1\cap N_2=\{e\}$.

For any element $(p,q)\in G$, $(p,q)=(p,1)\cdot (1,q)$, where $(p,1)\in N_1$ and $(1,q)\in N_2$ so it is true that $G=N_1N_2$.