proof of second isomorphism theorem for groups

First, we shall prove that $H K$ is a subgroup of $G$ : Since $e\in H$ and $e\in K$ , clearly $e=e^{2}\in HK$ . Take $h_{1},h_{2}\in H,k_{1},k_{2}\in K$ . Clearly $h_{1}k_{1},h_{2}k_{2}\in HK$ . Further,

$h_{1}k_{1}h_{2}k_{2}=h_{1}(h_{2}h_{2}^{-1})k_{1}h_{2}k_{2}=h_{1}h_{2}(h_{2}^{-% 1}k_{1}h_{2})k_{2}$

Since $K$ is a normal subgroup of $G$ and $h_{2}\in G$ , then $h_{2}^{-1}k_{1}h_{2}\in K$ . Therefore $h_{1}h_{2}(h_{2}^{-1}k_{1}h_{2})k_{2}\in HK$ , so $H K$ is closed under multiplication.

Also, $(hk)^{-1}\in HK$ for $h\in H$ , $k\in K$ , since

$(hk)^{-1}=k^{-1}h^{-1}=h^{-1}hk^{-1}h^{-1}$

and $hk^{-1}h^{-1}\in K$ since $K$ is a normal subgroup of $G$ . So $H K$ is closed under inverses, and is thus a subgroup of $G$ .

Since $H K$ is a subgroup of $G$ , the normality of $K$ in $H K$ follows immediately from the normality of $K$ in $G$ .

Clearly $H\cap K$ is a subgroup of $G$ , since it is the intersection of two subgroups of $G$ .

Finally, define $\phi\colon H\rightarrow HK/K$ by $\phi(h)=hK$ . We claim that $\phi$ is a surjective homomorphism from $H$ to $HK/K$ . Let $h_{0}k_{0}K$ be some element of $HK/K$ ; since $k_{0}\in K$ , then $h_{0}k_{0}K=h_{0}K$ , and $\phi(h_{0})=h_{0}K$ . Now

$\ker(\phi)=\{h\in H\mid\phi(h)=K\}=\{h\in H\mid hK=K\}$

and if $hK=K$ , then we must have $h\in K$ . So

$\ker(\phi)=\{h\in H\mid h\in K\}=H\cap K$

Thus, since $\phi(H)=HK/K$ and $\ker{\phi}=H\cap K$ , by the First Isomorphism Theorem we see that $H\cap K$ is normal in $H$ and that there is a canonical isomorphism between $H/(H\cap K)$ and $HK/K$ .

Title	proof of second isomorphism theorem for groups
Canonical name	ProofOfSecondIsomorphismTheoremForGroups
Date of creation	2013-03-22 12:49:47
Last modified on	2013-03-22 12:49:47
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	17
Author	yark (2760)
Entry type	Proof
Classification	msc 20A05
Related topic	ProofOfSecondIsomorphismTheoremForRings