2
$\begingroup$

Let $G$ be a group with identity element $e$. Let $H$ and $K$ be normal subgroup of $G$ such that $H∩K={e}$. Prove that
$(H×K)≅(H⊕K)$

$(H×K)$ is the internal direct product between $H$ and $K$
$(H⊕K)$ is the external direct product between $H$ and $K$

I know that since H and K be normal subgroup of G such that $H∩K={e}$, $hk=kh$ for all $h∈H$ and $k∈K$.
I also know that $(h,k)∈(H⊕K)$ . But I’m not sure I know how to link them together and show that $(H×K)≅(H⊕K)$

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Hint. Show that every element of the internal direct product may be uniquely represented as a product $hk$ for some $h\in H$, $k\in K$. Then use this to show that $\varphi(hk)=(h,k)$ is an isomorphism.

$\endgroup$
4
  • $\begingroup$ I have a theorem that say $H$ and $K$ be normal subgroup of $G$ such that $H∩K=e$, then $hk$ is unique for all $h∈H$ and $k∈K$ so I don't have to prove that. <br> Are you suggesting me to make a fucntion $φ:(H×K)→(H⊕K)$ defined by $φ(hk)=(h,k)$ then show this function is bijective and preserving structure? $\endgroup$
    – Diane Vanderwaif
    Apr 18, 2013 at 1:26
  • $\begingroup$ @DianeVanderwaif Yes, that's what an isomorphism is. And you should prove the theorem, don't just take it for granted, or you will not understand why it is true. (The proof is easy.) $\endgroup$
    – Alexander Gruber ♦
    Apr 18, 2013 at 1:44
  • $\begingroup$ ok, I proved that theorem, again. Now I'm going to prove it's injective. Just want to double check, if $(h,k)=(x,y)$ then $h=x$ and $k=y$. Is this true? $\endgroup$
    – Diane Vanderwaif
    Apr 18, 2013 at 3:22
  • $\begingroup$ very nice hint +1 $\endgroup$
    – Mikasa
    Apr 21, 2013 at 12:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .