# The order of the ring's product

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

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

Thanks

• 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. – drhab Jul 18 at 9:43

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 $$p^{6} = \Size{A} \ne \frac{\Size{A_{1}} \cdot \Size{A_{2}}}{\Size{A_{1} \cap A_{2}}} = \frac{p^{2} \cdot p^{3}}{p} = p^{4}.$$
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 $$|A_1||A_2|/|A_1\cap A_2|=4$$.
• If we define the 0 as the $F_2$'s identity, then your counterexample is correct. – se-hyuck yang Jul 18 at 10:41