In group case

Let the $A_1, A_2$ are subgroup of the A.

Then the order of the external product(or direct product)

$|A_1 \times A_2|$=$|A_1||A_2|$ holds.

On other hand, the order of the internal product(Block product) $|A_1 \cdot A_2|$ =$|A_1||A_2| \over |A_1\cap A_2|$ for $A_1 \cdot A_2 =\{a_1a_2| a_1 \in A_1 ,a_2 \in A_2 \}$ (As a point of the set's product view)

My question is

Do the above theorems still hold when we considering the ring case instead of the group case?

I mean Let's consider when the $A_1, A_2$ are rings. Though my guess one thing sure that unlike the group, the inner product of the rings are defined as

$A_1 \cdot A_2$= $\{\sum \ a_{1i}a_{2j}| a_{1i} \in A_1 ,a_{2j} \in A_2 \}$


  • $\begingroup$ The notation $A_1\times A_2$ is confusing here (see the answer of Wuestenfux). You can write $A_1A_2$ instead together with some explanation. Also the groups are apparantly subgroups of another group and you did not mention that. $\endgroup$ – drhab Jul 18 at 9:43

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$ $\DeclareMathOperator{\GF}{GF}$With your definition for the product of subrings (I take it that in your formula $$A_1 \cdot A_2 = \Set{\sum a_{1i}a_{2j} \mid a_{1i} \in A_1 ,a_{2j} \in A_2 }$$ $A_{1}, A_{2}$ are subrings of a common ring $A$), consider the following example.

Let $A = \GF(p^{6}),$ where $p$ is a prime, $A_{1} = \GF(p^{2}), A_{2} = \GF(p^{3})$, so that $A_{1} \cap A_{2} = \GF(p)$. As to $A_{1} \cdot A_{2}$, it is a subring of $A$ containing $A_{1}, A_{2}$, and thus $A_{1} \cdot A_{2} = A$. But $$p^{6} = \Size{A} \ne \frac{\Size{A_{1}} \cdot \Size{A_{2}}}{\Size{A_{1} \cap A_{2}}} = \frac{p^{2} \cdot p^{3}}{p} = p^{4}.$$


Does this count? I’m not sure since these two rings don’t share identity with the containing ring. But if admissible, it has the advantage of simplicity.

You can take the field $F_2$ of two elements, and the rings $A_1=F_2\times \{0\}$ and $A_2=\{0\}\times F_2$.

Then $|A_1\cdot A_2|=1$, but $|A_1||A_2|/|A_1\cap A_2|=4$.

  • $\begingroup$ If we define the 0 as the $F_2$'s identity, then your counterexample is correct. $\endgroup$ – se-hyuck yang Jul 18 at 10:41

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