Consider a (discrete or continuous) group $G$ and two normal/invariant subgroups $H_1$ and $H_2$ (possibly invariant/normal subgroups). Provided that $H_1,H_2\neq \{\operatorname{id}_G\}$, if $G/H_1\cong H_2$ and $G/H_2\cong H_1$, can we conclude that $G$ is the direct product of $H_1,H_2$, i.e. $G\cong H_1\times H_2$?
If not, then what can we conclude about the relationship between $G, H_1$ and $H_2$ if $G/H_1\cong H_2$ and $G/H_2\cong H_1$?
It's a good exercise to find counterexamples in small groups, not just for this question but for many basic group theory questions! Try cyclic groups, for example. If $n=ab$ is composite then $G=\Bbb Z_n$ has subgroups $H_1$ and $H_2$ of orders $a$ and $b$ respectively, then $G/H_1\cong H_2$ and $G/H_2\cong H_1$ since all subgroups and quotient groups of a cyclic group are cyclic (which are unique up to order).
(If you want $H_1\ne H_2$ of course choose $n$ with $a\ne b$.)
Likely we can't tell much at all from $G/H_1\cong H_2$ and $G/H_2\cong H_1$ because mere isomorphism is an equivalence relation with too much "freedom" to move around within. Even if $G=H\times K$ is a direct product, there can be an automorphism $\alpha$ for which $H_1=\alpha(H)$ and $H_2=K$ satisfy the same hypotheses but $G\ne H_1\times H_2$ is not a direct product!
For instance, suppose $H=\{(w,x,0,0)\}$ and $K=\{(0,0,y,z)\}$ within $G=\mathbb{Z}_2^4$ and let $\alpha$ be the automorphism (of $G$) defined by $\alpha(w,x,y,z)=(w,y,x,z)$ (it permutes the middle two coordinates); evidently $G=H\times K$ but defining $H_1=\alpha(H)=\{(w,0,x,0)\}$ and $H_2=K$ we know $G\ne H_1\times H_2$ since there is nontrivial intersection $H_1\cap H_2=\{(0,0,x,0)\}$! Yet $H\cong K\cong G/H\cong G/K$ all have the same isomorphism type as $H_1\cong H_2\cong G/H_1\cong G/H_2$.