• Oct 28th 2008, 07:41 AM
deniselim17
I have a theorem states that:
Let $G=H\times K$, and let $H_{1}\triangleleft H$ and $K_{1}\triangleleft K$. Then $H_{1}\times K_{1}\triangleleft G$ and $G/(H_{1}\times K_{1})\cong(H/H_{1})\times(K/K_{1})$.

Then I have this corollary:
If $G=H\times K$, then $G/(H\times\{{1}\})\cong K$.

Please check for me if my proof for the corollary is correct.

Proof:
We consider $H_{1}=H$, i.e. H has no proper normal subgroup.
Also,
$K_{1}=\{1\}$.
By last theorem, $G/(H_{1}\times K_{1})\cong (H/H_{1})\times(K/K_{1})$.
We have $G/(H\times\{1\})\cong (H/H)\times(K/\{1\})$.
Since H has no proper normal subgroup, so quotient group of H is not defined. Also, $(K/\{1\})=K$.
Hence, $G/(H\times\{1\})\cong K$
• Oct 28th 2008, 09:49 AM
ThePerfectHacker
Quote:

Originally Posted by deniselim17
Proof:
We consider $H_{1}=H$, i.e. H has no proper normal subgroup.
Also,
$K_{1}=\{1\}$.
By last theorem, $G/(H_{1}\times K_{1})\cong (H/H_{1})\times(K/K_{1})$.
We have $G/(H\times\{1\})\cong (H/H)\times(K/\{1\})$.
Since H has no proper normal subgroup, so quotient group of H is not defined. Also, $(K/\{1\})=K$.
Hence, $G/(H\times\{1\})\cong K$

It is correct except for a phrase you use "H has no proper normal subgroup", why not?
It does not matter. Just choose $H_1 = H$. We really do not care if it has proper normal subgroup (by the way it does, it has $\{ 1 \}$) and then apply the theorem and get your result. But then you say something that is again not necessary "so the quotient group of H is not defined", yes it is! Certainly $H\triangleleft H$ with $H/H \simeq \mathbb{Z}_1$. It is defined, it is just not a very interesting group.