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I was wondering if the following statement is true:

Let $G$ be a group with normal subgroups $H_1,H_2,...H_n$. Suppose $H_iH_j=G$ for all $i\neq j$. Then $G/H_1\cap H_2...\cap H_n\cong G/H_1 \times...\times G/H_n.$

There is a similar question here, which is the case when $n=2$. There is also a similar question here, but the conditions involve the index of a subgroup. I want to get rid of such condition so that the conclusion is applicable to some other situations.

But when I tried to proceed proof by induction to get the conclusion with arbitrary $n$, things became not so approachable.

If you think this is false, please give a counter-example. If you think this is true, please share your ideas of the proof. Thank you!

J. Wang
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1 Answers1

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This is false. For instance, let $G=(\mathbb{Z}/2\mathbb{Z})^2$ and let $H_1,H_2,H_3\subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $i\neq j$ but $G/(H_1\cap H_2\cap H_3)\cong G$ has $4$ elements while $G/H_1\times G/H_2\times G/H_3\cong(\mathbb{Z}/2\mathbb{Z})^3$ has $8$ elements.

Eric Wofsey
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  • Thank you. This example is clear. I will appreciate it if you can explain more why things fail to work for n>2. My intuitive understanding is that the information of a group is fully contained in the first two subgroup in the condition. So when we add one more, we count things redundantly. That’s why you and the other answer argue through the cardinality. However, is there a more in-depth reason why the statement fails? And how should we modify the condition to make it work? – J. Wang Oct 14 '18 at 00:38
  • I think you're asking the wrong question. The right question is why it does work for $n=2$. There are various proofs you can give in that case and you can see where they use the fact that $n=2$. – Eric Wofsey Oct 14 '18 at 00:41
  • Alternatively, it also does work if $G$ is a ring rather than just a group and the $H_i$ are ideals rather than just subgroups. The requirement that the $H_i$ are ideals is very strong, and in particular it implies that if $H_1+H_3=G$ and $H_2+H_3=G$ then $(H_1\cap H_2)+H_3=G$ as well, which allows you to make a proof by induction. – Eric Wofsey Oct 14 '18 at 00:46
  • If you're just working with groups, then the additional assumption you need is that $(H_1\cap\dots H_{i-1}\cap H_{i+1}\cap\dots H_n)H_i=G$ for all $i$. This means roughly that you can freely modify an element of $G$ modulo $H_i$ while not changing it modulo the other $H_j$'s, which is how you prove the natural homomorphism you want to be an isomorphism is surjective. – Eric Wofsey Oct 14 '18 at 00:52
  • Thanks a lot. Feel quite inspired now. – J. Wang Oct 14 '18 at 01:20
  • @EricWofsey could you elaborate on how $(H_1 \cap \cdots H_{i - 1} \cap H_{i + 1} \cap \cdots H_n)H_i = G$ for all $i$ makes the natural homomorphism surjective? I'm having trouble proving that – Raekye Dec 15 '18 at 15:13
  • @Raekye: Let $a\in G$. It suffices to find $b\in G$ which is $a$ mod $H_i$ but $1$ mod $H_j$ for all $j\neq i$, since the images of elements of this form are the subgroup $G/H_i$ of $G/H_1\times\dots G/H_n$ and these subgroups for all $i$ generate the whole product. Now by hypothesis, we can write $a=xy$ for $x\in H_1 \cap \cdots H_{i - 1} \cap H_{i + 1} \cap \cdots H_n$ and $y\in H_i$. But then $b=x$ is exactly the element we need: it is in every $H_j$ for $j\neq i$ but differs from $a$ by an element of $H_i$. – Eric Wofsey Dec 15 '18 at 17:37
  • @EricWofsey thank you, I see it now. I constructed the map going the other direction - if $H = H_1 \cap \cdots H_n$, and $K_i = H_1 \cap \cdots H_{i - 1} \cap H_{i + 1} \cap \cdots H_n$, we show $G/H_i \cong K_i/H$. Then define $\varphi \colon K_1/H \times \cdots K_n/H \to G/H$ by $(k_1H, \ldots k_nH) \mapsto k_1 \cdots k_nH$. This is analogue to the question here https://math.stackexchange.com/questions/254448/ . I proved that it's a well defined injective homomorphism, but I had trouble showing surjectivity this way. Do you know how this can be done (I want to try proving it both ways). – Raekye Dec 15 '18 at 21:11
  • @Raekye: Here's how it works for for $n=3$ (general $n$ is similar). We may assume $H$ is trivial (just mod it out everywhere). Given any $g\in G$, we write $g=ab$ for $a\in H_2\cap H_3$ and $b\in H_1$. Then we write $b=cd$ for $c\in H_1\cap H_3$ and $d\in H_2$. Notice then that $d\in H_1$ since $b,c\in H_1$. So we have $g=acd$ where $a\in K_1$, $c\in K_2$, and $d\in K_3$. – Eric Wofsey Dec 15 '18 at 21:53