The Third Isomorphism Theorem states that

Let $K$ and $H$ be two normal subgroups of group $G$ such that $K\leq H$. Then $G/H\cong (G/K)/(H/K)$.

Let $K_1$ and $K_2$ be two normal subgroups of $G$ such that $K_1,K_2\leq H$. Then shouldn't be the following be true? $$G/H\cong (G/K_1)/(H/K_2)$$ or $$G/H\cong (G/K_2)/(H/K_1)$$ Is the Third Isomorphism theorem just a generalisation of this?

Thanks in advance!


For $(G/K_1)/(H/K_2)$ to make sense, $H/K_2$ should be a subgroup of $G/K_1$, but there's no indication that that's the case. What if:

$G=\mathbb Z$
$H=2\mathbb Z$
$K_1=4\mathbb Z$
$K_2=6\mathbb Z$

Then there is no obvious relationship between $H/K_2$ and $G/K_1$. One is a group of order $3$ and the other is a group of order $4$, so neither group is isomorphic to a subgroup of the other.


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