Disprove that $C_6 \times C_{10} \times C_{12}=C_4\times C_{12}\times C_{15}$

Question

Disprove that $C_6 \times C_{10} \times C_{12}=C_4\times C_{12}\times C_{15}$ where $C_n$ denotes a cyclic group of order $n$.

By using the property that $(m,n)=1$ iff $C_m \times C_n=C_{mn}$, it can be shown that $$C_6 \times C_{10} \times C_{12}=C_2\times C_2 \times C_{12}\times C_{15}$$ And obviously $C_2\times C_2 \neq C_4$. But can I straight make the conclusion that $C_6 \times C_{10} \times C_{12}\neq C_4\times C_{12}\times C_{15}$ by just cancelling the cyclic factor $C_{12}$ and $C_{15}$?

Or asking in the other way, does $A\times B\cong C\times D$ and $A \cong C$ imply $B \cong D$?

And I know that I can disprove the question by finding some elements with order that does not exist in the other direct product but I can't find example.

Answer 1

Cancellations are tricky and one will be on slippery ground. Better avoid it. But what you have done gives a hint. Count the number of elements of order 2 on both sides. One can count 6 elements of order 2 on the left group, and just 3 on the right.

Answer 2

Use the property as much as you can to decompose and recompose factors:

$ A= C_6 \times C_{10} \times C_{12} \cong C_2 \times C_3 \times C_2 \times C_{5} \times C_{4} \times C_{3} \cong C_2 \times C_2 \times C_3 \times C_{3} \times C_{4} \times C_{5} \\ \quad\cong C_2 \times C_6 \times C_{60} $

$ B= C_4 \times C_{12} \times C_{15} \cong C_4 \times C_{3} \times C_{4} \times C_{3} \times C_{5} \cong C_3 \times C_{3} \times C_{4} \times C_{4} \times C_{5} \\ \quad\cong C_{12} \times C_{60} $

The number of elements of order $12$ in $A$ and $B$ are different and so $A\not\cong B$.