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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.

Thanks in advance!

Peter
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4 Answers4

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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.

Matt Samuel
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  • 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? – Peter Feb 23 '15 at 16:27
  • @Peter $f$ is surjective onto its image, which allows you to apply the isomorphism theorem to the image. – Matt Samuel Feb 23 '15 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? – Peter Feb 23 '15 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. – Matt Samuel Feb 23 '15 at 16:34
  • Thank you very much, I completely understand it now – Peter Feb 23 '15 at 16:45
  • @Peter You're welcome. – Matt Samuel Feb 23 '15 at 16:46
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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?

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

freedfromthereal
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$\textbf{Def:}$ 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).

Norman Contreras
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