Question about the proof of the third isomorphism thm (in Fraleigh).
Why is the $ker(\phi) = H$ in $\phi : G \rightarrow (G/K)/(H/K)$ by $\phi (a) = (aK)(H/K)$?
Where clearly $H, K \unlhd G$ and $K \unlhd H$.
Here's my argument.
I understand that all elements in the kernel map to $H/K$. Then, elts $a \in G$ mapping to $H/K$ must yield some coset $aK$ such that the element $(aK)(H/K) = H/K$. If $aK \subseteq H$, then certainly the inner product $(aK)(H) = H$. Then since $K \subseteq H$, any element $a \in H$ would make $(aK)(H) = H$. Then $H$ would be the Kernel.
First of all you have to be more careful with the notations. We can't have $aK \in H$, as the right one is a set too, so we must have $aK \subseteq H$. Secondly I don't see the need of assuming that $aK \in H$, why would you assume that?
Anyway a cleaner way would be to notice that $(aK)(H/K) = (H/K) \iff (aK) \in (H/K)$. But then as elements of $H/K$ are of the form $hK$ for some $h \in K$ we have that $(aK) \in (H/K) \iff a \in H$. From all this we can conclude that $\ker \phi = H$.