Question

Question: If ${\mathit{G}}_{\mathbf{1}}\mathbf{,}\mathbf{ }\mathbf{\dots }\mathbf{,}\mathbf{ }{\mathit{G}}_{\mathbf{n}}$are finite groups, prove that the order of $\left(a_1, a_2, \dots , a_n\right)\mathbf{}\mathit{i}\mathit{n}\mathbf{}{\mathit{G}}_{\mathbf{1}}\mathbf{\times }{\mathit{G}}_{\mathbf{2}}\mathbf{\times }\mathbf{\dots }\mathbf{\times }{\mathit{G}}_{\mathbf{n}}$in is the least common multiple of the orders$\left(\left|a_1\right|, \left|a_2\right|, \dots , \left|a_n\right|\right)$

Short answer

Answer

It is proved that, the order of $\left(a_1, a_2, \dots , a_n\right)\in G_1\times G_2\times \dots \times G_n$is$k=lcm\left(\left|a_1\right|, \left|a_2\right|, \dots , \left|a_n\right|\right)$ .

Step-by-step solution

Definition of an Element and Least Common Multiple

Definition of an Element:

Let G be a group and $a\in G$.

The smallest positive integer (if it exists) such that , is called order of .

Also, the result states that, if $a_m=e\left(=a^n\right)$for some positive integer , then .

Definition of Least Common Multiple (LCM)

The least common multiple of elements$a_1, a_2, \dots , a_n$ is defined as the least number, which is divisible by each$a_i, \forall i=1, 2, \dots , n$ .

To prove order of(a1, a2, …, an) divides K

$$\begin{gathered} \text{Let}\left(a_1, a_2, \dots , a_n\right)\in G_1\times G_2\times \dots \times G_n\text{and}\left|a_i\right|\text{isorderof}{a}_i, \forall i=1, 2, \dots , n\text{.} \\ \text{Let}k=lcm\left(\left|a_1\right|, \left|a_2\right|, \dots , \left|a_n\right|\right)\text{.} \\ \text{Then,}\left|a_i\right|\text{divides}\forall i=1, 2, \dots , n\text{.} \\ \text{Therefore,}{\left(a_1, a_2, \dots , a_n\right)}^k=\left(a_1^k, a_2^k, \dots , a_n^k\right)\text{.} \\ \text{Thisimpliesthat,}{\left(a_1, a_2, \dots , a_n\right)}^k=\left(e_{G_1}, e_{G_2}, \dots , e_{G_n}\right)\text{.} \\ \text{Hence,orderof}\left(a_1, a_2, \dots , a_n\right)\text{dividesK.} \end{gathered}$$

 Step 3: To prove order of (a1, a2, …, an)is K

$$\begin{gathered} \text{Now,letk'betheorderof}\left(a_1, a_2, \dots , a_n\right)\text{.} \\ \text{Then,}{\left(a_1, a_2, \dots , a_n\right)}^{k\text{'}}=\left(e_{G_1}, e_{G_2},\dots , e_{G_n}\right)\text{.} \\ \text{Thisimpliesthat,}\left(a_1^{k\text{'}}, a_2^{k\text{'}}, \dots , a_n^{k\text{'}}\right)=\left(e_{G_1}, e_{G_2}, \dots , e_{G_n}\right)\text{.} \\ \text{Thus,}k\text{'}\text{divides}\left|a_i\right|\text{,}\forall i=1, 2, \dots , n\text{.} \\ \text{Thisimpliesthatk'|kbydefinitionofLeastCommonMultiple.} \\ \text{Hence,theorderof}\left(a_1, a_2, \dots , a_n\right)\in G_1\times G_2\times \dots \times G_n\text{is}k=lcm\left(\left|a_1\right|, \left|a_2\right|, \dots , \left|a_n\right|\right)\text{.} \end{gathered}$$