Factor group of factor group and the 1st isomorphism theorem I have a question regarding the first isomorphism theorem:
Let $H,K$ be normal subgroups of $G_1$ and $G_2$. Can we then say:

$(G_1 \times G_2 )/(H\times {e_2})/(e_1\times K)\cong (G_1 \times G_2)/(H\times K)$

I have tried looking at this with the homomorphism theorem, but my brain cant compute so many fractions.
 A: Common abuse of notation: call the trivial subgroup of any group $1$. (This is in the context of abstract groups which are by default interpreted multiplicatively. In the event of, say, additive abelian groups we would use the notation $0$ instead.)
As Bernard points out in the comments, $1\times K$ is a subgroup of $G_1\times G_2$, not of the factor group $(G_1\times G_2)/(H\times 1)$, however the factor group has a subgroup corresponding to it.
The image of $1\times K$ under the projection map $G_1\times G_2\to (G_1\times G_2)/(H\times1)$ will be the normal subroup $(H\times K)/(H\times 1)$ of the factor group. When we quotient the factor group by it:
$$ \frac{(G_1\times G_2)/(H\times 1)}{(H\times K)/(H\times 1)}\cong \frac{G_1\times G_2}{H\times K}\cong \frac{G_1}{H}\times\frac{G_2}{K}. $$
The first is a special case of this: if $N\triangleleft M\triangleleft G$ are normal subroups, then we have
$$ \frac{G/N}{M/N}\cong \frac{G}{M} $$
with the isomorphism given by $gN(M/N)\leftrightarrow gN$. This applies in your context with the substitutions $G=G_1\times G_2$, $M=H\times K$ and $N=H\times 1$.
