Showing isomorphism between normal subgroups. This a fairly vague and basic example but i am struggling with it.
Suppose G is a group and it is the internal direct product of two subgroups H and K. $H\times K=G$ write down a explicit isomorphism $\phi :K \to G/H$ and prove that K is isomorphic to $G/H$.
Let $ k \in K $ then let $ \phi (k) = kH $ be the homomorphism.
Pf: consider $\phi (k_1) \phi (k_2) = k_1 H k_2 H $ since H is normal $k_1 H k_2 H 
 =k_1 k_2 H$ = $\phi ( k_1 k_2 ) $ so $ \phi $ is a homomorphism.
Show $ \phi $ is one to one
Assume $ \phi (k_1) = \phi (k_2)$ then $ k_1H=k_2H$ for some $h \in H $ then we have $ k_1h_1=k_2h_2$ or $ k^{-1}_{2}k_1=h_2h^{-1}_{1}$
Since $h_2h^{-1}_{1} \in H$ it implys that $k^{-1}_{2}k_1 \in H $ but $H \cap K = \{e\} $ so $k^{-1}_{2}k_1 = e$ or $k_1=k_2$ so one to one.
show $\phi $ is onto. I want to show that for all $a \in G/H $ that that the map $ \phi(k) \to kH $ hits every element of $G/H $.

$a\in G/H$ then $a= xH$, for $x\in G$. Note that $G = H\times K$, then there exists $h\in H$, $k\in K$ such that $x =kh$. So one has $\phi(k) = kh = kH = a$.
