# Prove: $G/(N_1 \cap N_2)$ is isomorphic with a subgroup of $(G/N_1) \times (G/N_2)$

I have to solve this exercise for my math study, but don't know how to do it. It's keeping me busy for 2 days now.

Let $G$ be a group and $N_{1}, N_{2}$ normal subgroups of $G$. Let $f: G \rightarrow (G/N_{1}) \times (G/N_{2})$ with $f(a) = (aN_{1}, aN_{2})$be a homomorphism with $Ker(f) = N_{1}\cap N_{2}$.

Prove that $G/N_{1}\cap N_{2}$ is isomorphic to a subgroup of $(G/N_{1}) \times (G/N_{1})$.

So far, I have done this:

$f$ is an homomorphism and $Ker(f) = N_{1}\cap N_{2}$, so $N_{1}\cap N_{2}$ is a normal subgroup of $G$. I also think that I have to use the fundamental homomorphism theorem, but I don't know how.

Could you please tell me how to prove this? I have really tried to solve this for a very long time, but I don't see what I have to do.

There are a lot of unnecessary distractions in this problem. In general, if $f:G\to H$ is a homomorphism with kernel $K$, then the image of $f$, which is a subgroup of $H$, is isomorphic to $G/K$. This follows from the first isomorphism theorem.

• Sorry, I forgot 1 thing in my question. I understand what you are saying, but that would mean that I have to prove that f is surjective, because then the image is equal to the range. But I think it is not possible to prove that. Or am I wrong? Feb 23, 2015 at 16:27
• @Peter $f$ is surjective onto its image, which allows you to apply the isomorphism theorem to the image. Feb 23, 2015 at 16:28
• But I don't know the image yet, so how is it possible to use the isomorphism theorem and say that the function is surjective? Feb 23, 2015 at 16:32
• @Peter it doesn't matter what the image is. Whatever the image, $f$ is surjective onto it. This follows from the definition of the image. $f$ doesn't need to be surjective onto the entire codomain. Feb 23, 2015 at 16:34
• Thank you very much, I completely understand it now Feb 23, 2015 at 16:45

When you say "Let $G$ be a group and $N_{1}, N_{2}$ normal subgroups of $G$. Let $f: G \rightarrow (G/N_{1}) \times (G/N_{1})$ be a homomorphism with $Ker(f) = N_{1}\cap N_{2}$", then you still have to define your homomorphism. This should be $f(g)=(gN_1, gN_2).$Then you have to prove that (a) this $f$ is a homomorphism and (b) that $ker(f)=N_1 \cap N_2$. Can you proceed from here?

• I am sorry. I have forgotten to add that. Now it's done Feb 23, 2015 at 16:27
• OK, then it is a matter of applying the First Isomorphism Theorem as noted by Matt Samuel. Feb 23, 2015 at 16:30
• Yes I understand that. But then I have to prove that the function f is surjective, because I don't know the image yet Feb 23, 2015 at 16:31
• No you do not have to prove surjectivity. For $G/ker(f) \cong im(f)$ and $im(f)$ is a subgroup of $G/N_1 \times G/N_2$. Feb 23, 2015 at 18:31

If you prove that the map $f$ is surjective, then you have the claim. Indeed, by the first isomorphism theorem you have that $Im(f) \cong (G/N_1) \times (G/N_2)$.

• Yes, that's absolutely true, but how is it possible to prove that this function is surjective? Feb 23, 2015 at 16:29
• Is the map $f$ from $G$ to $G/N_1 \times G/N_2$ or to $G/N_1 \times G/N_1$? Feb 23, 2015 at 16:40
• Oh sorry, you are right Feb 23, 2015 at 16:44

$$\textbf{Def:}$$ Let $$A,B$$ be groups. Then, $$A \times B=\{(a,b)|a \in A, b \in B\}$$. $$A \times B$$ is a group by defining $$(a_1,b_1) \cdot (a_2,b_2) \,\, \text{for} \,\, a_1,a_2 \in A \,\, \text{and} \,\, b_1,b_2 \in B$$ called the $$\textbf{Direct Product}$$ of $$A$$ and $$B$$. (You can check that this is a group by checking closure, identity, associativity, and inverse).

$$\textbf{Proof:}$$ Define $$\varphi:G \rightarrow G/N_1 \times G/N_2$$ by $$\varphi(G)=(N_1g,N_2g) \,\, \forall g \in G$$.
So, $$\text{ker}(\varphi)=\{g \in G | \varphi(G)=1_{G/N_1 \times G/N_2}\}=\{g \in G | \varphi(G)=(N_1,N_2)\}=\{g \in G | (N_1g,N_2g)=(N_1,N_2)\}$$.
We know that $$N_1g=N_1$$ if and only if $$g \in N$$.
Thus, $$\text{ker}(\varphi)=\{g \in G | g \in N_1 and g \in N_2\} = \{ g \in G | g \in N_1 \cap N_2\} = N_1 \cap N_2$$.
Then, by the First Isomorphism Theorem, $$G/N_1 \cap N_2 \cong \varphi(G)$$. $$\,\, \blacksquare$$
(To be more rigorous, you can show that $$\varphi$$ is a homomorphism).