please check for me

**deniselim17** · October 28th 2008, 06:41 AM

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, $G/(H\times\{1\})\cong K$

**ThePerfectHacker** · October 28th 2008, 08:49 AM

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.

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