Generalization of Chinese remainder theorem to non-normal subgroups I am trying to see if the following generalization holds or if there are counter-examples.
Let $H, K<G$ be subgroups of $G$ (not necessarily normal). It is not too hard to prove that
$$\phi: G\to G/H\times G/K; g\mapsto (gH, gK)$$
is surjective iff $HK = G$.
I wonder if you can generalize this to the case of $n$ subgroups. Let $H_1,\dots,H_n$ be subgroups of $G$. The map
$$\phi: G\to G/H_1\times \dots\times G/H_n; g\mapsto (gH_1,\cdots, gH_n)$$
is surjective iff for $i\neq j$, $H_iH_j = G$.
I can see reasons that this may not hold, then a counter-example would be appreciated.
This is in the same spirit as the Chinese remainder theorem in ring theory, but there we have more structures to help us.
 A: Let $G$ be the Klein $4$-group, $C_2\times C_2$, with generators $x$ and $y$ (so the elements of $G$ are $(1,1)$, $(x,1)$, $(1,y)$, and $(x,y)$, with $x^2=y^2=1$).
Let $H_1 = C_2\times \{1\}$, $H_2 = \{e\}\times C_2$, and $H_3 = \{(1,1),(x,y)\}$.  Then $H_iH_j=G$ for $i\neq j$.  For simplicity, denote the cosets of $H_i$ by $H_i = (1,1)H_i$ and $M_i = G\setminus H_i$.
I claim that no element of $G$ maps to the element $(M_1,M_2,M_3)$ of $(G/H_1)\times (G/H_2) \times (G/H_3)$.
Indeed, if $g\in G$ mapped to $(M_1,M_2,M_3)$, then we would have that $g\notin H_1$, $g\notin H_2$, and $g\notin H_3$. But $H_1\cup H_2\cup H_3 = G$, so this is impossible.  Or you can note that $(G/H_1)\times(G/H_2)\times (G/H_3)$ has eight elements, so there can be no surjection from the group of order $4$.
Thus, the map $\phi$ is not surjective in this case. Note that here we even have the $H_i$ normal in $G$.
More generally, if $G$ is the set-theoretic union of proper subgroups $\{H_i\}_{i\in I}$, then the corresponding map $\phi$ cannot be surjective since no element of $G$ can have an image that is not $H_i$ in the $i$th coordinate for all $i$, even if we had that $H_iH_j=G$ for all $i\neq j$.
A: Thanks to the above answer by Arturo Magidin, I think the most natural generalization of the Chinese remainder theorem from rings with identity to groups are as follows.
Ring version: Let $A$ be a ring and ${\mathfrak a}_1,\dots,{\mathfrak a}_n$ ideals of $A$. The homomorphism
$$
\phi: A\to \prod_{i=1}^n (A/{\mathfrak a}_i);\ x\mapsto (x+{\mathfrak a}_1, \dots, x+{\mathfrak a}_n)$$
is surjective iff
$${\mathfrak a_i}+ \bigg(\bigcap_{j\neq i} {\mathfrak a}_j\bigg) = (1),\quad\forall 1\leq i\leq n.$$
Group version: Let $G$ be a group and $H_1,\dots,H_n$ normal subgroups of $G$. The homomorphism
$$
\psi: G\to \prod_{i=1}^n (G/H_i);\ g\mapsto (gH_1, \dots, gH_n)$$
is surjective iff
$$H_i \cdot \bigg(\bigcap_{j\neq i} H_j\bigg) = G,\quad\forall 1\leq i\leq n.$$
The proofs are not hard.
