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this question is similar to another question, but that question uses the second isomorphism theorem that I am not at liberty to use. The question is this: Showing that $G/(H\cap K)\cong (G/H)\times (G/K)$. In this question, it is also assumed that $G=HK$, which I cannot assume.

Suppose that G is a group and H and K are normal subgroups of G. By considering the map $\theta : G \rightarrow G/H \times G/K$ defined by $g\rightarrow (gH)(gK)$, show that $G/(H\cap K)$ is isomorphic to a subgroup of $G/H \times G/K$.

I have attempted to tackle the question like so

$|G/(H\cap K)|=\frac{|G|}{|H\cap K|}$.

$|G/H \times G/K|=\frac{|G/H||G/K|}{| G/H \cap G/K |}=\frac{|G||G|}{|H||K|}\frac{1}{| H \cap K |}=|G/(H\cap K)|\frac{G}{|H||K|}. $

So we have that need a subgroup of $|G/H \times G/K|$ that works. I was wondering if I take the subgroup such that $H \cap K =1$, would this provide what I needed, and if so how would I define the required isomorphism.

Thanks!

Siddharth Panigrahi
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Catherine Drysdale
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2 Answers2

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I think you are assuming all your groups are finite. This works for infinite groups as well.

Do you know the first isomorphism theorem? All you need to do is to show this map: $g\to (gH,gK)$ is well defined (trivial) and that the kernel is $H \cap K$. Then the first isomorphism theorem proves that $G/H\cap K \cong Im(\phi)$, and the image of a homomorphism is a subgroup, which proves it.

CPM
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  • Thank you so much - you answe is very clear! – Catherine Drysdale Jun 01 '17 at 08:38
  • Glad it helped. Don't forget to prove it is a homomorphism also. This is also very straightforward. Show $\phi(g_1 g_2)=\phi(g_1)\phi(g_2)$. – CPM Jun 01 '17 at 08:40
  • the next part of the question is suppose that $G$ is an abelian group of order 66. Show that $G$ is a cyclic group and show that tehre are proper groups $H$ and $K$ of $G$ such that $G \cong G/H \times G/K$. How would you go about this? I know it is cheeky to ask. – Catherine Drysdale Jun 01 '17 at 08:50
  • This is a different kind of question, so you really should start a new question thread in case someone in the future has the same type of problem. Just mention there that you think you should use the result from this one. – CPM Jun 01 '17 at 08:54
  • Okay, I will do this! – Catherine Drysdale Jun 01 '17 at 08:56
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Unfortunately I do not know how to proceed with your combinatorial proof.

Assuming you are allowed the first isomorphism theorem( and even if you have not seen it yet it is easy to prove), what is the kernel of $\theta$? Well $\theta(g)=(eH,eK)$ implies that $gH=H$ and $gK =K$. Hence $g\in H\cap K$. It is easy to see the other inclusion, so $\ker(\theta)=H\cap K$, and then apply the first isomorphism theorem.